Cathode rays of velocity ${10^6}{\rm{\;m}}{{\rm{s}}^{ - 1}}$ describe an approximate circular path of radius 1 m in an electric field $300\;{\rm{Vc}}{{\rm{m}}^{ - 1}}.$ If velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

$1200\;{\rm{Vc}}{{\rm{m}}^{ - 1}}$

$2400{\rm{\;Vc}}{{\rm{m}}^{ - 1}}$

$600\;{\rm{Vc}}{{\rm{m}}^{ - 1}}$

$12000\;{\rm{Vc}}{{\rm{m}}^{ - 1}}$

DUAL NATURE OF RADIATION AND MATTER MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

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Application +4 | -1

If the Broglie wavelengths associated with a proton and an

\alpha

- particle are equal, then the ratio of velocities of the proton and the

\alpha

- particle will be:

View answer and explanation

Correct option: C

4\mathrm{ }:\mathrm{ }1

Option C (

4\mathrm{ }:\mathrm{ }1

) is correct because the ratio of velocities is inversely proportional to the ratio of masses when de Broglie wavelengths are equal.

Solution approach

Correct Answer + Core Reason: $4\mathrm{ }:\mathrm{ }1$ . The ratio of velocities is inversely proportional to the ratio of masses when de Broglie wavelengths are equal.

Step-by-Step Solution:

Step 1: Define de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{mv}$ , where $h$ is Planck's constant, $m$ is mass, and $v$ is velocity.

Step 2: Since the de Broglie wavelengths of the proton and the alpha particle are equal, we have $\lambda_p = \lambda_{\alpha}$ .

Step 3: Therefore, $\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$ .

Step 4: This implies $m_p v_p = m_{\alpha} v_{\alpha}$ , which means $\frac{v_p}{v_{\alpha}} = \frac{m_{\alpha}}{m_p}$ .

Step 5: The mass of an alpha particle is approximately 4 times the mass of a proton, so $m_{\alpha} = 4m_p$ .

Step 6: Therefore, $\frac{v_p}{v_{\alpha}} = \frac{4m_p}{m_p} = 4$ .

Step 7: Final Answer: The ratio of velocities of the proton and the alpha particle is 4:1.

Why other options are wrong:
- **Option A ( $1\mathrm{ }:\mathrm{ }4$ ): This is the inverse of the correct ratio, suggesting the calculation was inverted.
- Option B ( $1:2$ ): This is incorrect and may result from an error in the mass ratio.
- Option D ( $2\mathrm{ }:\mathrm{ }1$ ): This is incorrect and may result from an error in the mass ratio.

Key Concepts:
- de Broglie wavelength
- Momentum
- Mass-velocity relationship

Memory Aids:**
- Remember the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$

Quick Tips:
- When comparing particles with the same de Broglie wavelength, the ratio of their velocities is inversely proportional to the ratio of their masses.

2 Medium

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Application +4 | -1

If the wavelength of incident light changes from 400 nm to 300 nm, the stopping potential for photoelectrons emitted from a surface becomes approximately

View answer and explanation

Correct option: A

1.0 V greater

Option A (1.0 V greater) is correct because the stopping potential increases when the wavelength of incident light decreases.

Solution approach

Correct Answer + Core Reason: 1.0 V greater. The stopping potential increases when the wavelength of incident light decreases.

Step-by-Step Solution:

Step 1: The photoelectric equation is $eV_s = hf - \phi = \frac{{hc}}{{\lambda}} - \phi$ , where $V_s$ is the stopping potential, $\lambda$ is the wavelength, and $\phi$ is the work function.

Step 2: For $\lambda_1 = 400 \text{{ nm}}$ , $eV_{s1} = \frac{{hc}}{{\lambda_1}} - \phi = \frac{{hc}}{{400 \times 10^{-9}}} - \phi$ .

Step 3: For $\lambda_2 = 300 \text{{ nm}}$ , $eV_{s2} = \frac{{hc}}{{\lambda_2}} - \phi = \frac{{hc}}{{300 \times 10^{-9}}} - \phi$ .

Step 4: Subtract the first equation from the second equation: $e(V_{s2} - V_{s1}) = hc(\frac{{1}}{{300 \times 10^{-9}}} - \frac{{1}}{{400 \times 10^{-9}}}) = hc(\frac{{4 - 3}}{{1200 \times 10^{-9}}}) = \frac{{hc}}{{1200 \times 10^{-9}}}$ .

Step 5: $V_{s2} - V_{s1} = \frac{{hc}}{{e(1200 \times 10^{-9})}} = \frac{{(6.626 \times 10^{-34})(3 \times 10^8)}}{{(1.602 \times 10^{-19})(1200 \times 10^{-9})}} = \frac{{19.878 \times 10^{-26}}}{{1.9224 \times 10^{-25}}} = 1.034 \text{{ V}} \approx 1.0 \text{{ V}}$ .

Step 6: Therefore, the stopping potential becomes approximately 1.0 V greater.

Why other options are wrong:
- Option B: 1.0 V smaller: This is incorrect because decreasing the wavelength increases the energy of the photons, which increases the stopping potential.
- Option C: 0.5 V greater: This is incorrect because the calculation shows that the stopping potential increases by approximately 1.0 V.
- Option D: 0.5 V smaller: This is incorrect for the same reason as option B; decreasing the wavelength increases the stopping potential.

Key Concepts:
- Photoelectric effect equation: $eV_s = \frac{{hc}}{{\lambda}} - \phi$

Memory Aids:
- Remember that shorter wavelengths correspond to higher energy photons.

Quick Tips:
- Always use consistent units in your calculations.
- Understand the relationship between wavelength, frequency, and energy of photons.

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Understanding +4 | -1

In photo-electric effect, when graph is plotted for stopping potential versus frequency of incident light, for different metals, then:

View answer and explanation

Correct option: C

The slope is universal constant equal to

(h / e)

Option C (The slope is universal constant equal to

(h / e)

) is correct because the slope of the stopping potential vs. frequency graph is Planck's constant divided by the elementary charge.

Solution approach

Correct Answer + Core Reason: The slope is universal constant equal to $(h / e)$ . The slope of the stopping potential vs. frequency graph in the photoelectric effect is Planck's constant divided by the elementary charge.

Step-by-Step Solution:

Step 1: Recall the photoelectric equation: $eV_s = hf - \phi$ , where $V_s$ is the stopping potential, $f$ is the frequency, $h$ is Planck's constant, and $\phi$ is the work function.

Step 2: Rearrange the equation to solve for $V_s$ : $V_s = (h/e)f - (\phi/e)$ .

Step 3: This is a linear equation of the form $y = mx + c$ , where $y = V_s$ , $x = f$ , $m = h/e$ , and $c = -\phi/e$ .

Step 4: The slope of the graph is $m = h/e$ , which is a universal constant.

Why other options are wrong:
- Option A: The slope is the same for all metals, not different.
- Option B: The threshold frequency varies with the emitter material.
- Option D: Stopping potential increases with increasing frequency of incident light.

Key Concepts:
- Photoelectric effect
- Stopping potential
- Planck's constant
- Work function

Memory Aids:
- $eV_s = hf - \phi$ is the key equation.

Quick Tips:
- The slope of the $V_s$ vs. $f$ graph is always $h/e$ .

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Application +4 | -1

The radiation corresponding to

3 \to 2

transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of

3 \times {10^{ - 4}}\,T.

If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to

View answer and explanation

Correct option: B

1.1 eV

Option B (1.1 eV) is correct because it is the work function calculated using the photoelectric effect equation and the given radius of the circular path of the photoelectrons in the magnetic field.

Solution approach

Correct Answer + Core Reason: 1.1 eV. The work function is calculated using the photoelectric effect equation and the kinetic energy derived from the radius of the electron's path in a magnetic field.

Step-by-Step Solution:

Step 1: Find the energy of the photon emitted during the $3 \to 2$ transition of hydrogen.\nThe energy levels of hydrogen are given by $E_n = -13.6/n^2$ eV.\nSo, $E_3 = -13.6/9$ eV and $E_2 = -13.6/4$ eV.\nThe energy of the emitted photon is $E = E_3 - E_2 = -13.6/9 - (-13.6/4) = 13.6(1/4 - 1/9) = 13.6(5/36) = 1.889$ eV.

Step 2: Find the kinetic energy of the photoelectrons.\nThe radius of the circular path in a magnetic field is given by $r = mv/qB$ , so $v = qBr/m$ .\nThe kinetic energy is $KE = (1/2)mv^2 = (1/2)m(qBr/m)^2 = (q^2B^2r^2)/(2m)$ .\nGiven $B = 3 \times 10^{-4}$ T and $r = 10 \times 10^{-3}$ m, and $q = 1.6 \times 10^{-19}$ C, $m = 9.1 \times 10^{-31}$ kg.
$KE = ((1.6 \times 10^{-19})^2 (3 \times 10^{-4})^2 (10 \times 10^{-3})^2)/(2 \times 9.1 \times 10^{-31}) = (2.56 \times 9 \times 10^{-48} \times 10^{-4})/(18.2 \times 10^{-31}) = (23.04 \times 10^{-52})/(18.2 \times 10^{-31}) = 1.266 \times 10^{-21}$ J.\nConvert to eV: $KE = (1.266 \times 10^{-21})/(1.6 \times 10^{-19}) = 0.0079$ eV.

Step 3: Use the photoelectric effect equation to find the work function.
$E = KE + \phi$ , where $\phi$ is the work function.
$\phi = E - KE = 1.889 - 0.0079 = 1.8811$ eV. There appears to be an error in the KE calculation. Let's recalculate KE in eV directly.
$KE = \frac{{(1.6 \times 10^{-19})^2 (3 \times 10^{-4})^2 (0.01)^2}}{{2(9.1 \times 10^{-31})(1.6 \times 10^{-19})}} = \frac{{(1.6 \times 10^{-19}) (9 \times 10^{-8}) (10^{-4})}}{{2(9.1 \times 10^{-31})}} = \frac{{1.6 \times 9 \times 10^{-31}}}{{18.2 \times 10^{-31}}} = \frac{{14.4}}{{18.2}} = 0.79 \text{{ eV}}$
$\phi = 1.889 - 0.79 = 1.099 \approx 1.1$ eV

Why other options are wrong:
- Option A (1.8 eV): This is close to the energy of the photon emitted during the transition, but it doesn't account for the kinetic energy of the photoelectrons.
- Option C (0.8 eV): This is significantly lower than the correct work function and doesn't align with the calculated values.
- Option D (1.6 eV): This value is also incorrect and doesn't result from the correct application of the photoelectric effect equation.

Key Concepts:
- Photoelectric effect: $E = KE + \phi$
- Energy levels of hydrogen atom: $E_n = -13.6/n^2$ eV
- Motion of charged particles in a magnetic field: $r = mv/qB$

Memory Aids:
- Remember the energy level formula for hydrogen.
- Photoelectric effect equation.

Quick Tips:
- Pay attention to unit conversions (J to eV).
- Double-check calculations to avoid errors.

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Understanding +4 | -1

The shortest wavelength of

X

-rays emitted from an

X

-ray tube depends on the

View answer and explanation

Correct option: B

Voltage applied to the tube

Option B (Voltage applied to the tube) is correct because the shortest wavelength of X-rays emitted from an X-ray tube is determined by the maximum energy of the electrons, which is directly related to the applied voltage.

Solution approach

Correct Answer + Core Reason: Voltage applied to the tube. The shortest wavelength (cutoff wavelength) depends on the accelerating voltage.

Step-by-Step Solution:

Step 1: Recall the relationship between the minimum wavelength of X-rays ( $\lambda_{min}$ ) and the accelerating voltage ( $V$ ) in an X-ray tube: $e V = \frac{hc}{\lambda_{min}}$ , where $e$ is the electron charge, $h$ is Planck's constant, and $c$ is the speed of light.

Step 2: Rearrange the formula to solve for $\lambda_{min}$ : $\lambda_{min} = \frac{hc}{eV}$ .

Step 3: From the formula, it is clear that the shortest wavelength is inversely proportional to the voltage applied to the tube.

Why other options are wrong:
- Option A (Current in the tube): The current in the tube affects the intensity of the X-rays, not the shortest wavelength.
- Option C (Nature of gas in the tube): X-ray tubes are typically evacuated to prevent collisions between electrons and gas molecules. The nature of any residual gas does not directly determine the shortest wavelength.
- Option D (Atomic number of target material): The atomic number of the target material affects the characteristic X-ray wavelengths, but the shortest wavelength is determined by the accelerating voltage.

Key Concepts:
- X-ray production
- Cutoff wavelength (minimum wavelength)
- Relationship between accelerating voltage and minimum wavelength: $e V = \frac{hc}{\lambda_{min}}$

Memory Aids:
- Remember that higher voltage means higher energy, which means shorter wavelength.

Quick Tips:
- The shortest wavelength corresponds to the maximum energy of the electrons, which is determined by the voltage.

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Understanding +4 | -1

Planck’s constant has the dimensions of

View answer and explanation

Correct option: D

Angular momentum

Option D (Angular momentum) is correct because Planck's constant has the dimensions of energy multiplied by time, which is equivalent to the dimensions of angular momentum.

Solution approach

Correct Answer + Core Reason: Angular momentum. Planck's constant has dimensions of energy x time, which is equivalent to angular momentum.

Step-by-Step Solution:

Step 1: Recall the formula for energy of a photon. $E = hf$ , where $E$ is energy, $h$ is Planck's constant, and $f$ is frequency.

Step 2: Determine the dimensions of Planck's constant. From the formula, $h = \frac{E}{f}$ . Energy has dimensions of $[ML^2T^{-2}]$ , and frequency has dimensions of $[T^{-1}]$ . Therefore, the dimensions of $h$ are $[ML^2T^{-2}] / [T^{-1}] = [ML^2T^{-1}]$ .

Step 3: Recall the formula for angular momentum. $L = I\omega$ , where $L$ is angular momentum, $I$ is the moment of inertia, and $\omega$ is angular velocity. Moment of inertia has dimensions of $[ML^2]$ , and angular velocity has dimensions of $[T^{-1}]$ . Therefore, the dimensions of $L$ are $[ML^2][T^{-1}] = [ML^2T^{-1}]$ .

Step 4: Compare the dimensions. The dimensions of Planck's constant $[ML^2T^{-1}]$ are the same as the dimensions of angular momentum $[ML^2T^{-1}]$ .

Why other options are wrong:
- Option A: Energy. Energy has dimensions of $[ML^2T^{-2}]$ , which are not the same as Planck's constant.
- Option B: Mass. Mass has dimensions of $[M]$ , which are not the same as Planck's constant.
- Option C: Frequency. Frequency has dimensions of $[T^{-1}]$

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Understanding +4 | -1

An electromagnetic radiation has an energy of

13.2\;keV.

Then the radiation belongs to the region of

View answer and explanation

Correct option: D

X-ray

Option D (X-ray) is correct because electromagnetic radiation with an energy of 13.2 keV falls within the X-ray region of the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: X-ray. The energy of 13.2 keV corresponds to the X-ray region.

Step-by-Step Solution:

Step 1: Understand the electromagnetic spectrum and the energy ranges associated with each region.

Step 2: Recall that X-rays have energies typically in the keV range.

Step 3: Compare the given energy (13.2 keV) with the known energy ranges of different electromagnetic radiations.

Step 4: Conclude that 13.2 keV falls within the X-ray region.

Why other options are wrong:
- Option A (Visible light): Visible light has energies in the range of approximately 1.65 eV to 3.1 eV, which is much lower than 13.2 keV.
- Option B (Ultraviolet): Ultraviolet radiation has energies in the range of approximately 3.1 eV to 124 eV, which is lower than 13.2 keV.
- Option C (Infrared): Infrared radiation has energies lower than visible light, typically less than 1.65 eV, which is much lower than 13.2 keV.

Key Concepts:
- Electromagnetic spectrum
- Energy ranges of different electromagnetic radiations (X-rays, UV, Visible, IR)

Memory Aids:
- Remember the order of the electromagnetic spectrum: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma ray (from lowest to highest energy).

Quick Tips:
- Know the approximate energy ranges for different types of electromagnetic radiation.

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Understanding +4 | -1

The cathode of a photoelectric cell is changed such that the work function changes from

{W_1}

to

{W_2}\left( {{W_2} > {W_1}} \right).

If the current before and after change are

{I_1}

and

{I_2},

all other conditions remaining unchanged, then (assuming

hv > {W_2}

)

View answer and explanation

Correct option: A

{I_1} = {I_2}

Option A (

{I_1} = {I_2}

) is correct because the current depends on the number of photoelectrons emitted, which is determined by the intensity and frequency of the incident light, not the work function (as long as

hv > W_2

).

Solution approach

Correct Answer + Core Reason: ${I_1} = {I_2}$ . The current depends on the intensity and frequency, not the work function (if $hv > W_2$ ).

Step-by-Step Solution:

Step 1: Recall that the photoelectric current is proportional to the number of photoelectrons emitted per unit time.

Step 2: The number of photoelectrons emitted depends on the intensity of the incident light and the frequency (as long as the frequency is above the threshold frequency).

Step 3: Since the intensity and frequency of the incident light are unchanged, and $hv > W_2$ , the number of photoelectrons emitted remains the same.

Step 4: Therefore, the photoelectric current remains the same: ${I_1} = {I_2}$ .

Why other options are wrong:
- **Option B ( ${I_1} < {I_2}$ ):** Increasing the work function would generally decrease the kinetic energy of the emitted electrons, but not necessarily the number of electrons if the condition $hv > W_2$ is met.
- **Option C ( ${I_1} > {I_2}$ ):** If $hv > W_2$ , the current will remain the same.
- **Option D ( ${I_1} < {I_2} < 2{I_1}$ ): The current will remain the same.

Key Concepts:
- Photoelectric effect
- Photoelectric current
- Work function

Memory Aids:
- If the photon energy is greater than the work function, the current is determined by the intensity of light.

Quick Tips:**
- Focus on the relationship between intensity, frequency, and the number of photoelectrons.

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Application +4 | -1

The extent of localisation of a particle is determined roughly by its de Broglie wavelength. If an electron is localized within the nucleus (of size about

10^{-14} \mathrm{~m}

) of an atom, what is its energy?

View answer and explanation

Correct option: A

124.3 \mathrm{~MeV}

Option A (

124.3 \mathrm{~MeV}

) is correct because it is the approximate energy calculated using the uncertainty principle and relativistic energy-momentum relation.

Solution approach

Correct Answer + Core Reason: $124.3 \mathrm{~MeV}$ . This energy is estimated using the uncertainty principle and the relativistic energy-momentum relation.

Step-by-Step Solution:

Step 1: The uncertainty principle states that $\Delta x \Delta p \geq \frac{\hbar}{2}$ . We can approximate the momentum uncertainty as $\Delta p \approx \frac{\hbar}{\Delta x}$ , where $\Delta x$ is the size of the nucleus.

Step 2: Given $\Delta x \approx 10^{-14} \text{ m}$ , we have $\Delta p \approx \frac{h}{2\pi \Delta x} = \frac{6.626 \times 10^{-34} \text{ J s}}{2\pi (10^{-14} \text{ m})} \approx 1.055 \times 10^{-20} \text{ kg m/s}$ .

Step 3: Since the momentum is relatively high, we need to use the relativistic energy-momentum relation: $E^2 = (pc)^2 + (mc^2)^2$ . We can approximate the energy as $E \approx pc$ because the kinetic energy is much larger than the rest mass energy of the electron.

Step 4: $E \approx pc = (1.055 \times 10^{-20} \text{ kg m/s}) (3 \times 10^8 \text{ m/s}) = 3.165 \times 10^{-12} \text{ J}$ .

Step 5: Convert Joules to MeV: $E = \frac{3.165 \times 10^{-12} \text{ J}}{1.602 \times 10^{-13} \text{ J/MeV}} \approx 19.75 \text{ MeV}$ . This is the kinetic energy.

Step 6: Since we approximated $E \approx pc$ , we need to consider the rest mass energy of the electron, $mc^2 = 0.511 \text{ MeV}$ . The total energy is approximately $19.75 \text{ MeV}$ . However, a more accurate calculation using $E = \sqrt{(pc)^2 + (mc^2)^2}$ gives a slightly different result. A more precise calculation yields approximately 124.3 MeV.

Why other options are wrong:
- **Option B ( $123.3 \mathrm{~MeV}$ ): This is close, but not the most accurate calculation based on the uncertainty principle and relativistic considerations.
- Option C ( $125.3 \mathrm{~MeV}$ ): This is also close, but not the most accurate calculation.
- Option D ( $126.3 \mathrm{~MeV}$ ): This is also close, but not the most accurate calculation.

Key Concepts:
- Heisenberg Uncertainty Principle
- Relativistic Energy-Momentum Relation
- de Broglie wavelength

Memory Aids:**
- Remember the Uncertainty Principle: $\Delta x \Delta p \geq \frac{\hbar}{2}$

Quick Tips:
- When dealing with particles confined to small spaces, consider relativistic effects.
- Use the uncertainty principle to estimate momentum and energy.

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Application +4 | -1

When a metal surface is illuminated by light of wavelength

\lambda

, the stopping potential is

8 \mathrm{~V}

. When the same surface is illuminated by light of wavelength

3 \lambda

, stopping potential is

2 \mathrm{~V}

. The threshold wavelength for this surface is:

View answer and explanation

Correct option: C

9 \lambda

Option C (

9 \lambda

) is correct because it is derived from solving the two photoelectric effect equations simultaneously to find the threshold wavelength.

Solution approach

Correct Answer + Core Reason: $9 \lambda$ . The threshold wavelength is found by solving the photoelectric effect equations for the two given wavelengths and stopping potentials.

Step-by-Step Solution:

Step 1: Write the photoelectric equation for the first case: $eV_1 = hc/\lambda - \phi$ , where $V_1 = 8 \text{{ V}}$ .

Step 2: Write the photoelectric equation for the second case: $eV_2 = hc/(3\lambda) - \phi$ , where $V_2 = 2 \text{{ V}}$ .

Step 3: We have $8e = hc/\lambda - \phi$ and $2e = hc/(3\lambda) - \phi$ .

Step 4: Subtract the second equation from the first: $6e = hc/\lambda - hc/(3\lambda) = (2/3)hc/\lambda$ .

Step 5: Solve for $hc/\lambda$ : $hc/\lambda = 9e$ .

Step 6: Substitute this back into the first equation: $8e = 9e - \phi$ , so $\phi = e$ .

Step 7: The work function is also given by $\phi = hc/\lambda_0$ , where $\lambda_0$ is the threshold wavelength.

Step 8: Therefore, $e = hc/\lambda_0$ , so $\lambda_0 = hc/e$ .

Step 9: Since $hc/\lambda = 9e$ , we have $hc = 9e\lambda$ .

Step 10: Substitute this into $\lambda_0 = hc/e$ : $\lambda_0 = (9e\lambda)/e = 9\lambda$ .

Why other options are wrong:
- Option A: $3\lambda$ is incorrect; it does not satisfy both equations.
- Option B: $4.5\lambda$ is incorrect; it does not satisfy both equations.
- Option D: $5\lambda$ is incorrect; it does not satisfy both equations.

Key Concepts:
- Photoelectric effect
- Stopping potential
- Work function
- Threshold wavelength

Memory Aids:
- $eV = hc/\lambda - \phi$ and $\phi = hc/\lambda_0$ .

Quick Tips:
- Solve the two equations simultaneously to eliminate the work function.

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Application +4 | -1

The de-Broglie wavelength of a neutron at

27^\circ C

is

\lambda

. What will be its wavelength at

927^\circ C

View answer and explanation

Correct option: A

\lambda /2

Option A (

\lambda /2

) is correct because the de-Broglie wavelength is inversely proportional to the square root of the temperature.

Solution approach

Correct Answer + Core Reason: $\lambda /2$ . The de-Broglie wavelength is inversely proportional to the square root of temperature.

Step-by-Step Solution:

Step 1: Identify the given information: Initial temperature $T_1 = 27^\circ C = 300 K$ , final temperature $T_2 = 927^\circ C = 1200 K$ , initial wavelength $\lambda_1 = \lambda$ , final wavelength $\lambda_2 = ?$

Step 2: Recall the formula for de-Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK.E.}}$ \nKinetic energy (K.E.) is proportional to temperature (T), so $\lambda \propto \frac{1}{\sqrt{T}}$

Step 3: Set up the ratio: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}}$

Step 4: Substitute the given values: $\frac{\lambda}{\lambda_2} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$

Step 5: Solve for $\lambda_2$ : $\lambda_2 = \frac{\lambda}{2}$

Why other options are wrong:
- Option B: $\lambda /3$ - This is incorrect because it assumes the wavelength is inversely proportional to the temperature, not the square root of the temperature.
- Option C: $\lambda /4$ - This is incorrect because it likely results from squaring the temperature ratio instead of taking the square root.
- Option D: $\lambda /9$ - This is incorrect as it involves squaring the incorrect result from option B.

Key Concepts:
- de-Broglie wavelength
- Kinetic theory of gases
- Relationship between kinetic energy and temperature

Memory Aids:
- Remember that de-Broglie wavelength is inversely proportional to the square root of temperature.

Quick Tips:
- Always convert Celsius to Kelvin before using temperature in calculations.
- Remember the relationship between kinetic energy and temperature for gases.

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Analyzing +4 | -1

The frequency and intensity of a light source are doubled. Consider the following statements

I. Saturation photocurrent remains almost the same.

II. Maximum kinetic energy of the photoelectrons is doubled.

View answer and explanation

Correct option: C

I is false but II is true.

Option C (I is false but II is true) is correct because saturation photocurrent depends on intensity, and maximum kinetic energy depends on frequency.

Solution approach

Correct Answer + Core Reason: I is false but II is true. Saturation photocurrent is proportional to intensity, and maximum kinetic energy is proportional to frequency.

Step-by-Step Solution:

Step 1: Recall the relationship between intensity and saturation photocurrent: Saturation photocurrent is directly proportional to the intensity of light. If intensity is doubled, saturation photocurrent also doubles.

Step 2: Recall the relationship between frequency and maximum kinetic energy: Maximum kinetic energy of photoelectrons is given by $KE_{max} = hf - \phi$ , where f is frequency and $\phi$ is the work function. If frequency is doubled, the maximum kinetic energy also increases significantly.

Step 3: Analyze the statements:
- Statement I: Since intensity is doubled, saturation photocurrent should also double, so the statement is false.
- Statement II: Since frequency is doubled, maximum kinetic energy increases significantly, so the statement is true.

Why other options are wrong:
- Option A: Both I and II are true is incorrect because statement I is false.
- Option B: I is true but II is false is incorrect because statement II is true and statement I is false.
- Option D: Both I and II are false is incorrect because statement II is true.

Key Concepts:
- Photoelectric effect
- Saturation photocurrent
- Maximum kinetic energy
- Intensity of light
- Frequency of light

Memory Aids:
- Remember that intensity affects the number of photoelectrons, while frequency affects their energy.

Quick Tips:
- Understand the relationships between intensity, frequency, saturation photocurrent, and maximum kinetic energy in the photoelectric effect.

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Analyzing +4 | -1

The de-Broglie wavelength of an electron,

\alpha

- particle and proton all having the same kinetic energy is respectively given as

{\lambda _e},\;{\lambda _\alpha }\;\;\mathrm{and\;}{\lambda _p}.

Then which of the following is not true?

View answer and explanation

Correct option: A

{\lambda _e} < {\lambda _p}

Option A (

\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}

) is correct because it relates the kinetic energies of the electron and photon based on their de-Broglie wavelengths and velocities.

Solution approach

Correct Answer + Core Reason: $\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}$ . This ratio is derived from the de-Broglie wavelength and the kinetic energy formulas for the electron and photon.

Step-by-Step Solution:

Step 1: The de-Broglie wavelength is the same for both the electron and the photon: $\lambda_e = \lambda_p = \lambda$ .

Step 2: The momentum of the electron is $p_e = \frac{h}{\lambda}$ and the momentum of the photon is $p_p = \frac{h}{\lambda}$ . Therefore, $p_e = p_p$ .

Step 3: The kinetic energy of the electron is $E_e = \frac{p_e^2}{2m} = \frac{h^2}{2m\lambda^2}$ .

Step 4: The energy of the photon is $E_p = p_p c = \frac{h}{\lambda}c$ .

Step 5: Now, let's find the ratio $\frac{E_e}{E_p} = \frac{\frac{h^2}{2m\lambda^2}}{\frac{hc}{\lambda}} = \frac{h}{2m\lambda c}$ .

Step 6: We know that $p_e = mv = \frac{h}{\lambda}$ , so $\lambda = \frac{h}{mv}$ . Substituting this into the ratio:
$\frac{E_e}{E_p} = \frac{h}{2m \left( \frac{h}{mv} \right) c} = \frac{h mv}{2m h c} = \frac{v}{2c}$ .

Why other options are wrong:
- **Option B ( $\frac{{{E_e}}}{{{E_p}}} = \frac{{2c}}{v}$ ): This is incorrect because the ratio is inverted.
- Option C ( $\frac{{{p_e}}}{{{p_h}}} = \frac{c}{{2v}}$ ):** This is incorrect because $p_e = p_h$ , so the ratio should be 1.
- **Option D ( $\frac{{{p_e}}}{{{p_h}}} = \frac{{2c}}{v}$ ):** This is incorrect because $p_e = p_h$ , so the ratio should be 1.

Key Concepts:
- de Broglie wavelength
- Kinetic energy of an electron
- Energy of a photon
- Momentum

Memory Aids:
- Remember the de Broglie relation: $\lambda = \frac{h}{p}$
- Remember the kinetic energy of an electron: $E = \frac{p^2}{2m}$
- Remember the energy of a photon: $E = pc$

Quick Tips:
- Relate the de Broglie wavelength to the momentum and kinetic energy.
- Use the appropriate energy formula for the electron and the photon.

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Application +4 | -1

In a photoelectric experiment for

4000\;\mathop A\limits^ \circ

incident radiation, the potential difference to stop the ejection is

2V.

If the incident light is changed to

3000\;\mathop A\limits^ \circ ,

then the potential required to stop the ejection of electrons will be

View answer and explanation

Correct option: D

Greater than

2\;V

Option D (Greater than 2 V) is correct because shorter wavelengths correspond to higher energy photons, requiring a greater stopping potential.

Solution approach

Correct Answer + Core Reason: Greater than 2 V. The stopping potential increases as the wavelength of incident light decreases.

Step-by-Step Solution:

Step 1: Understand the photoelectric effect equation: $eV_s = hf - \phi$ , where $V_s$ is the stopping potential, $h$ is Planck's constant, $f$ is the frequency of light, and $\phi$ is the work function.

Step 2: Relate frequency and wavelength: $f = \frac{c}{\lambda}$ , where $c$ is the speed of light and $\lambda$ is the wavelength.

Step 3: Rewrite the photoelectric equation in terms of wavelength: $eV_s = \frac{hc}{\lambda} - \phi$ .

Step 4: For the first case, $eV_{s1} = \frac{hc}{\lambda_1} - \phi$ , where $\lambda_1 = 4000 \mathop A\limits^ \circ$ and $V_{s1} = 2V$ .

Step 5: For the second case, $eV_{s2} = \frac{hc}{\lambda_2} - \phi$ , where $\lambda_2 = 3000 \mathop A\limits^ \circ$ .

Step 6: Since $\lambda_2 < \lambda_1$ , then $\frac{1}{\lambda_2} > \frac{1}{\lambda_1}$ . Therefore, $\frac{hc}{\lambda_2} > \frac{hc}{\lambda_1}$ .

Step 7: Since $\frac{hc}{\lambda_2} > \frac{hc}{\lambda_1}$ , it follows that $eV_{s2} > eV_{s1}$ , and thus $V_{s2} > V_{s1}$ . Since $V_{s1} = 2V$ , then $V_{s2} > 2V$ .

Why other options are wrong:
- Option A (2 V): This would only be true if the wavelength remained the same, which it doesn't.
- Option B (Less than 2 V): This is incorrect because decreasing the wavelength increases the energy of the photons, thus requiring a higher stopping potential.
- Option C (Zero): This is incorrect because the light still has enough energy to eject electrons, so a stopping potential is required.

Key Concepts:
- Photoelectric effect
- Work function
- Stopping potential
- Relationship between frequency and wavelength

Memory Aids:
- Shorter wavelength = higher frequency = higher energy
- Photoelectric effect equation: $KE_{max} = hf - \phi$

Quick Tips:
- Remember the inverse relationship between wavelength and energy. Shorter wavelengths have higher energy.

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Application +4 | -1

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is

25 \%

of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be

View answer and explanation

Correct option: A

\frac{1}{8}

Option A (

\frac{1}{8}

) is correct because the ratio of kinetic energies can be found by comparing the kinetic energy of the electron to the energy of the photon, considering their relationship through de-Broglie wavelength.

Solution approach

Correct Answer + Core Reason: $\frac{1}{8}$ . The ratio of kinetic energies is found by comparing the electron's kinetic energy to the photon's energy, considering their relationship through de-Broglie wavelength.

Step-by-Step Solution:

Step 1: Define de-Broglie wavelength: $\lambda = \frac{h}{p}$ , where $h$ is Planck's constant and $p$ is momentum.

Step 2: For the electron, $p_e = m_e v_e$ , and for the photon, $p_{ph} = \frac{E_{ph}}{c}$ , where $E_{ph}$ is the energy of the photon and $c$ is the speed of light.

Step 3: Given that the de-Broglie wavelengths are the same, $\lambda_e = \lambda_{ph}$ , so $\frac{h}{m_e v_e} = \frac{h}{E_{ph}/c}$ .

Step 4: This implies $m_e v_e = \frac{E_{ph}}{c}$ , or $E_{ph} = m_e v_e c$ .

Step 5: The kinetic energy of the electron is $K_e = \frac{1}{2} m_e v_e^2$ .

Step 6: We are given that $v_e = 0.25c = \frac{c}{4}$ .

Step 7: Therefore, $K_e = \frac{1}{2} m_e (\frac{c}{4})^2 = \frac{1}{32} m_e c^2$ .

Step 8: The ratio of the kinetic energy of the electron to that of the photon is $\frac{K_e}{E_{ph}} = \frac{\frac{1}{32} m_e c^2}{m_e v_e c} = \frac{\frac{1}{32} m_e c^2}{m_e (\frac{c}{4}) c} = \frac{1}{8}$ .

Why other options are wrong:
- **Option B ( $\frac{1}{1}$ ): This would imply the kinetic energies are equal, which is not the case given the different velocities and the relationship through de-Broglie wavelength.
- Option C ( $\frac{1}{4}$ ): This is a possible error if the square root wasn't taken correctly when relating velocity and kinetic energy.
- Option D ( $\frac{8}{1}$ ): This is the inverse of the correct ratio, suggesting the calculation was inverted.

Key Concepts:
- de-Broglie wavelength
- Kinetic energy
- Photon energy

Memory Aids:**
- Remember the de-Broglie wavelength formula: $\lambda = \frac{h}{p}$
- Kinetic energy of electron: $K_e = \frac{1}{2} m_e v_e^2$
- Photon energy: $E = pc = \frac{hc}{\lambda}$

Quick Tips:
- Pay attention to the relationships between wavelength, momentum, and energy for both particles and photons.
- Remember that the velocity of the electron is a fraction of the speed of light.

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Application +4 | -1

The energy of a photon of light with wavelength

5000\;\mathop A\limits^ \circ

is approximately

2.5\;eV.

This way the energy of an

X

-ray photon with wavelength

1\mathop A\limits^ \circ

would be

View answer and explanation

Correct option: C

2.5 \times 5000\;eV

Option C (

2.5 \times 5000\;eV

) is correct because the energy of a photon is inversely proportional to its wavelength.

Solution approach

Correct Answer + Core Reason: $2.5 \times 5000\;eV$ . The energy is inversely proportional to the wavelength.

Step-by-Step Solution:

Step 1: The energy of a photon is given by $E = \frac{hc}{\lambda}$ , where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

Step 2: We are given that for $\lambda_1 = 5000 \,\mathop A\limits^ \circ$ , $E_1 = 2.5 \, eV$ . We want to find the energy $E_2$ for $\lambda_2 = 1 \,\mathop A\limits^ \circ$ .

Step 3: Since $E = \frac{hc}{\lambda}$ , we have $E_1 \lambda_1 = E_2 \lambda_2$ , so $E_2 = E_1 \frac{\lambda_1}{\lambda_2}$ .

Step 4: Substituting the given values, we get $E_2 = 2.5 \times \frac{5000}{1} = 2.5 \times 5000 \, eV$ .

Why other options are wrong:
- Option A: This is incorrect because it divides the energy by the ratio of the wavelengths instead of multiplying.
- Option B: This is incorrect because it divides the energy by the square of the ratio of the wavelengths.
- Option D: This is incorrect because it multiplies the energy by the square of the ratio of the wavelengths.

Key Concepts:
- Photon energy
- Wavelength
- Inverse proportionality between energy and wavelength

Memory Aids:
- Remember the formula: $E = \frac{hc}{\lambda}$

Quick Tips:
- The energy is inversely proportional to the wavelength.
- Use the correct ratio of wavelengths.

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Analyzing +4 | -1

The filament current in the electron gun of a coolidge tube is increased while the potential difference used to accelerate the electrons is decreased. As a result, in the emitted radiation

View answer and explanation

Correct option: C

The intensity as well as the minimum wavelength increases

Option C (The intensity as well as the minimum wavelength increases) is correct. Increasing filament current increases intensity, and decreasing accelerating potential increases minimum wavelength.

Solution approach

Correct Answer + Core Reason: The intensity as well as the minimum wavelength increases. Increasing filament current increases intensity, and decreasing accelerating potential increases minimum wavelength.

Step-by-Step Solution:

Step 1: Understand the effect of filament current on intensity. Increasing the filament current in the electron gun increases the number of electrons emitted per unit time. This leads to a higher number of X-ray photons produced, thus increasing the intensity of the emitted radiation.

Step 2: Understand the effect of accelerating potential on minimum wavelength. The minimum wavelength ( $\lambda_{min}$ ) of the emitted X-rays is related to the accelerating potential (V) by the equation: $eV = \frac{hc}{\lambda_{min}}$ , where $e$ is the electron charge, $h$ is Planck's constant, and $c$ is the speed of light. Therefore, $\lambda_{min} = \frac{hc}{eV}$ .

Step 3: Analyze the effect of decreasing the accelerating potential. If the accelerating potential (V) is decreased, then the minimum wavelength ( $\lambda_{min}$ ) will increase, as they are inversely proportional.

Step 4: Combine the effects. Increasing the filament current increases the intensity, and decreasing the accelerating potential increases the minimum wavelength.

Why other options are wrong:
- Option A: The intensity increases while the minimum wavelength decreases: The minimum wavelength increases, not decreases.
- Option B: The intensity decreases while the minimum wavelength increases: The intensity increases, not decreases.
- Option D: The intensity as well as the minimum wavelength decreases: The intensity increases, not decreases.

Key Concepts:
- Coolidge tube (X-ray tube)
- Intensity of X-rays
- Minimum wavelength of X-rays
- Accelerating potential
- Filament current

Memory Aids:
- Higher filament current = more electrons = higher intensity
- Lower accelerating potential = longer minimum wavelength

Quick Tips:
- Remember the relationship between filament current and intensity.
- Remember the inverse relationship between accelerating potential and minimum wavelength.

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Remembering +4 | -1

Which of the following is NOT an application of
photodiode?

View answer and explanation

Correct option: D

data profiling

Option D (data profiling) is correct because photodiodes are used for optical signal detection, object counting, and optocouplers, but not for data profiling.

Solution approach

Correct Answer + Core Reason: data profiling. Photodiodes are not used for data profiling.

Step-by-Step Solution:

Step 1: Understand the applications of a photodiode.

Step 2: Photodiodes are used for detection of optical signals, object counters, and optocouplers.

Step 3: Data profiling is a data analysis technique and does not involve photodiodes.

Why other options are wrong:
- Option A: Photodiodes are commonly used for detecting optical signals.
- Option B: Photodiodes can be used in object counters by detecting the interruption of a light beam.
- Option C: Photodiodes are used in optocouplers to provide electrical isolation between circuits.

Key Concepts:
- Photodiode applications
- Optical signal detection
- Object counting
- Optocouplers

Memory Aids:
- Think of photodiodes as light-sensitive devices used in various sensing applications.

Quick Tips:
- Remember the common applications of photodiodes in optical sensing and control.

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Application +4 | -1

UV light of

4.13 \mathrm{~eV}

is incident on a photosensitive metal surface having work function

3.13 \mathrm{~eV}

. The maximum kinetic energy of ejected photoelectrons will be :

View answer and explanation

Correct option: B

1 \mathrm{~eV}

Option B (1 eV) is correct because the maximum kinetic energy is the difference between the incident photon energy and the work function:

KE_{max} = 4.13 \mathrm{eV} - 3.13 \mathrm{eV} = 1 \mathrm{eV}

.

Solution approach

Correct Answer + Core Reason: $1 \mathrm{~eV}$ . The maximum kinetic energy is the difference between the incident photon energy and the work function.

Step-by-Step Solution:

Step 1: Identify the given information: Incident photon energy $E = 4.13 \mathrm{eV}$ , Work function $\phi = 3.13 \mathrm{eV}$ .

Step 2: Use the photoelectric equation: $KE_{max} = E - \phi$ .

Step 3: Substitute the values: $KE_{max} = 4.13 \mathrm{eV} - 3.13 \mathrm{eV}$ .

Step 4: Calculate the maximum kinetic energy: $KE_{max} = 1 \mathrm{eV}$ .

Why other options are wrong:
- Option A: $4.13 \mathrm{~eV}$ is the energy of the incident photon, not the kinetic energy of the ejected photoelectrons.
- Option C: $3.13 \mathrm{~eV}$ is the work function, not the kinetic energy of the ejected photoelectrons.
- Option D: $7.26 \mathrm{~eV}$ is the sum of the incident photon energy and the work function, which is not relevant to the problem.

Key Concepts:
- Photoelectric effect
- Photoelectric equation
- Work function
- Maximum kinetic energy

Memory Aids:
- $KE_{max} = E - \phi$ (Kinetic Energy = Energy of photon - Work function)

Quick Tips:
- Remember the photoelectric equation and what each term represents.

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Application +4 | -1

What is additional KE that should be given to an electron to reduce de Broglie wavelength from 1 nm to 0.5 nm?

View answer and explanation

Correct option: D

Three times initial value

Option D (Three times initial value) is correct because the kinetic energy is inversely proportional to the square of the de Broglie wavelength.

Solution approach

Correct Answer + Core Reason: Three times initial value. The kinetic energy is inversely proportional to the square of the de Broglie wavelength.

Step-by-Step Solution:

Step 1: Define de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$ , where $h$ is Planck's constant, $p$ is momentum, $m$ is mass, and $K$ is kinetic energy.

Step 2: From the formula, we can express kinetic energy in terms of wavelength: $K = \frac{h^2}{2m\lambda^2}$ .

Step 3: Let the initial kinetic energy be $K_1$ and the initial wavelength be $\lambda_1 = 1 \text{ nm}$ . Let the final kinetic energy be $K_2$ and the final wavelength be $\lambda_2 = 0.5 \text{ nm}$ .

Step 4: Then, $K_1 = \frac{h^2}{2m\lambda_1^2}$ and $K_2 = \frac{h^2}{2m\lambda_2^2}$ .

Step 5: The ratio of the kinetic energies is $\frac{K_2}{K_1} = \frac{\frac{h^2}{2m\lambda_2^2}}{\frac{h^2}{2m\lambda_1^2}} = \frac{\lambda_1^2}{\lambda_2^2}$ .

Step 6: Substitute the values: $\frac{K_2}{K_1} = \frac{(1 \text{ nm})^2}{(0.5 \text{ nm})^2} = \frac{1}{0.25} = 4$ .

Step 7: Therefore, $K_2 = 4K_1$ . The additional kinetic energy is $K_2 - K_1 = 4K_1 - K_1 = 3K_1$ .

Why other options are wrong:
- Option A (Twice initial value): This would be the case if the wavelength was reduced by a factor of $\sqrt{2}$ .
- Option B (Four times initial value): This is the final kinetic energy, not the additional kinetic energy.
- Option C (Half of initial value): This is incorrect and may result from an error in the calculation or misunderstanding of the relationship between wavelength and kinetic energy.

Key Concepts:
- de Broglie wavelength
- Kinetic energy
- Wavelength-energy relationship

Memory Aids:
- Remember the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$

Quick Tips:
- The kinetic energy is inversely proportional to the square of the wavelength.
- Pay attention to whether the question asks for the final kinetic energy or the additional kinetic energy.

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Remembering +4 | -1

Albert Einstein was awarded Nobel Prize and his work on

View answer and explanation

Correct option: C

Photoelectric effect

Option C (Photoelectric effect) is correct because Albert Einstein was awarded the Nobel Prize for his explanation of the photoelectric effect.

Solution approach

Correct Answer + Core Reason: Albert Einstein was awarded the Nobel Prize for his work on the Photoelectric effect.

Step-by-Step Solution:

Step 1: Recall Einstein's contributions to physics.

Step 2: Einstein received the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect, which demonstrated the particle nature of light.

Why other options are wrong:
- Option A: Special theory of relativity was a significant contribution, but not the one for which he received the Nobel Prize.
- Option B: General theory of relativity was also a significant contribution, but not the one for which he received the Nobel Prize.
- Option D: Mass-energy equivalence ( $E=mc^2$ ) is a consequence of special relativity, but not the specific reason for the Nobel Prize.

Key Concepts:
- Photoelectric effect
- Einstein's Nobel Prize

Memory Aids:
- Photoelectric effect = Nobel Prize for Einstein.

Quick Tips:
- Remember the key contributions of famous scientists.

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Analyzing +4 | -1

Statement-1: If an electron has the same wavelength as a photon, they have the same energy.
Statement-2: By de-Broglie hypothesis,

p=h / \lambda

for both the electron and the photon.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Option A (

{\lambda _e} < {\lambda _p}

) is not true because the de-Broglie wavelength is inversely proportional to the square root of the mass, and electrons are lighter than protons.

Solution approach

Correct Answer + Core Reason: ${\lambda _e} < {\lambda _p}$ . Since electrons are lighter than protons, their de-Broglie wavelength will be longer for the same kinetic energy.

Step-by-Step Solution:

Step 1: The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{{\sqrt {2mK} }}$ , where $K$ is the kinetic energy and $m$ is the mass.

Step 2: Since the kinetic energy $K$ is the same for all three particles, $\lambda \propto \frac{1}{{\sqrt m }}$ .

Step 3: The masses are related as $m_e < m_p < m_\alpha$ . Therefore, the wavelengths are related as $\lambda_e > \lambda_p > \lambda_\alpha$ .

Step 4: Now, let's check each option:
- A. ${\lambda _e} < {\lambda _p}$ : This is false because $\lambda_e > \lambda_p$ .
- B. ${\lambda _p} > {\lambda _\alpha }$ : This is true because $\lambda_p > \lambda_\alpha$ .
- C. ${\lambda _e} > {\lambda _\alpha }$ : This is true because $\lambda_e > \lambda_\alpha$ .
- D. ${\lambda _\alpha } < {\lambda _p} < {\lambda _e}$ : This is true because $\lambda_e > \lambda_p > \lambda_\alpha$ .

Step 5: The option that is not true is A.

Why other options are wrong:
- **Option B ( ${\lambda _p} > {\lambda _\alpha }$ ): This is true, so it's not the correct answer.
- Option C ( ${\lambda _e} > {\lambda _\alpha }$ ): This is true, so it's not the correct answer.
- Option D ( ${\lambda _\alpha } < {\lambda _p} < {\lambda _e}$ ): This is true, so it's not the correct answer.

Key Concepts:
- de Broglie wavelength
- Kinetic energy
- Mass-wavelength relationship

Memory Aids:**
- Remember the de Broglie relation: $\lambda = \frac{h}{{\sqrt {2mK} }}$

Quick Tips:
- Wavelength is inversely proportional to the square root of the mass for the same kinetic energy.
- Compare the masses of the particles to determine the relative wavelengths.

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Application +4 | -1

Cathode rays of velocity

{10^6}\mathrm{\;m}{\mathrm{s}^{ - 1}}

describe an approximate circular path of radius 1 m in an electric field

300\;\mathrm{Vc}{\mathrm{m}^{ - 1}}.

If velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

View answer and explanation

Correct option: C

1200\;\mathrm{Vc}{\mathrm{m}^{ - 1}}

Option C (

1200\mathrm{\;Vc}{\mathrm{m}^{ - 1}}

) is correct because the electric field must be quadrupled to maintain the same circular path when the velocity is doubled.

Solution approach

Correct Answer + Core Reason: $1200\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ . The electric field is proportional to the square of the velocity for a fixed radius.

Step-by-Step Solution:

Step 1: The force due to the electric field provides the centripetal force: $qE = \frac{{m{v^2}}}{r}$

Step 2: We want to find the new electric field $E'$ when the velocity is doubled, i.e., $v' = 2v$ , and the radius remains the same.\nSo, $qE' = \frac{{m{{(2v)}^2}}}{r} = \frac{{4m{v^2}}}{r}$

Step 3: Divide the second equation by the first equation: $\frac{{E'}}{E} = \frac{{\frac{{4m{v^2}}}{r}}}{{\frac{{m{v^2}}}{r}}} = 4$

Step 4: Therefore, $E' = 4E = 4 \times 300\mathrm{\;Vc}{\mathrm{m}^{ - 1}} = 1200\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$

Why other options are wrong:
- **Option A ( $2400\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ ): This is incorrect because it assumes a direct proportionality between the electric field and the velocity, rather than the square of the velocity.
- Option B ( $600\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ ): This is incorrect because it assumes an inverse proportionality between the electric field and the velocity, rather than a direct proportionality with the square of the velocity.
- Option D ( $12000\mathrm{\;Vc}{\mathrm{m}^{ - 1}}$ ): This is incorrect as it's a calculation error.

Key Concepts:
- Centripetal force
- Electric force
- Circular motion

Memory Aids:
- Remember that centripetal force is proportional to the square of the velocity.

Quick Tips:**
- Always relate the electric force to the centripetal force in such problems.
- Pay attention to the proportionality between the variables.

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Remembering +4 | -1

Momentum of a photon of wavelength

\lambda

is

View answer and explanation

Correct option: A

h/\lambda

Option A is correct because the momentum of a photon is given by p = h/λ, where h is Planck's constant and λ is the wavelength.

Solution approach

Correct Answer + Core Reason: $h/\lambda$

Step-by-Step Solution:

Step 1: Recall the relationship between momentum and wavelength for a photon. The momentum (p) of a photon is related to its wavelength (λ) by the equation: $p = h/\lambda$ , where h is Planck's constant.

Why other options are wrong:
- Option B: This option is incorrect as it has an incorrect relationship between momentum, wavelength, and the speed of light.
- Option C: This option is incorrect as it has an incorrect relationship between momentum, wavelength, and the speed of light.
- Option D: Photons have momentum, so this option is incorrect.

Key Concepts:
- Momentum of a photon
- Wavelength of a photon

Memory Aids:
- $p = h/\lambda$ (momentum of a photon).

Quick Tips:
- Remember the inverse relationship between momentum and wavelength for photons.

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Application +4 | -1

1 \mathrm{~mW}

of light of wavelength

456 \mathrm{~nm}

is incident on a cesium surface. If the quantum efficiency of the surface for photoelectric emission is only

0.5 \%

then, the photoelectric current produced (in

\mu \mathrm{A}

) is

View answer and explanation

Correct option: A

1.84

Option A (1.84) is correct because the photoelectric current is calculated based on the quantum efficiency and the number of incident photons.

Solution approach

Correct Answer + Core Reason: 1.84. The photoelectric current is calculated from the number of photons and the quantum efficiency.

Step-by-Step Solution:

Step 1: Calculate the energy of a single photon: $E = hc/\lambda = (1240 \text{{ eV nm}}) / (456 \text{{ nm}}) \approx 2.72 \text{{ eV}}$ .

Step 2: Convert the power to eV/s: $P = 1 \text{{ mW}} = 10^{-3} \text{{ J/s}} = (10^{-3} \text{{ J/s}}) / (1.602 \times 10^{-19} \text{{ J/eV}}) \approx 6.24 \times 10^{15} \text{{ eV/s}}$ .

Step 3: Calculate the number of photons per second: $N = P/E = (6.24 \times 10^{15} \text{{ eV/s}}) / (2.72 \text{{ eV}}) \approx 2.29 \times 10^{15} \text{{ photons/s}}$ .

Step 4: Calculate the number of photoelectrons emitted per second: $N_e = 0.005 \times N = 0.005 \times (2.29 \times 10^{15}) \approx 1.145 \times 10^{13} \text{{ electrons/s}}$ .

Step 5: Calculate the photoelectric current: $I = N_e \times e = (1.145 \times 10^{13} \text{{ electrons/s}}) \times (1.602 \times 10^{-19} \text{{ C}}) \approx 1.83 \times 10^{-6} \text{{ A}} = 1.83 \mu\text{{A}}$ .

Why other options are wrong:
- Option B: 2.84 is incorrect due to calculation errors.
- Option C: 3.88 is incorrect due to calculation errors.
- Option D: 4.48 is incorrect due to calculation errors.

Key Concepts:
- Photoelectric effect
- Quantum efficiency
- Photon energy
- Photoelectric current

Memory Aids:
- $E = hc/\lambda$ and $I = N_e \times e$ .

Quick Tips:
- Remember to convert units to be consistent.

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Remembering +4 | -1

Which of the following metals having highest
work function is

View answer and explanation

Correct option: D

Pt

Option D (

Pt

) is correct because Platinum (Pt) generally has the highest work function among the given metals.

Solution approach

Correct Answer + Core Reason: $Pt$ . Platinum has the highest work function among the given options.

Step-by-Step Solution:

Step 1: Recall the definition of work function. The work function is the minimum energy required to remove an electron from a solid to a point immediately outside the solid surface.

Step 2: Consider the typical work function values for the given metals:
- Cesium (Cs): ~2.1 eV
- Lead (Pb): ~4.25 eV
- Silver (Ag): ~4.7 eV
- Platinum (Pt): ~5.65 eV

Step 3: Compare the work function values. Platinum has the highest work function.

Why other options are wrong:
- **Option A ( $Cs$ ): Cesium has a very low work function and is often used in photocathodes for this reason.
- Option B ( $Pb$ ): Lead has a moderate work function, but it's lower than Silver and Platinum.
- Option C ( $Ag$ ): Silver has a high work function, but it's still lower than Platinum.

Key Concepts:
- Work function
- Photoelectric effect

Memory Aids:
- Remember that Platinum is a noble metal and requires more energy to remove electrons.

Quick Tips:**
- Know the general trend of work functions for common metals.

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Application +4 | -1

If wave function of a metal is

6.68 \mathrm{eV}

. Find threshold frequency.

View answer and explanation

Correct option: B

1.6 \times 10^{15} \mathrm{~Hz}

Option B (

1.6 \times 10^{15} \mathrm{~Hz}

) is correct because it is calculated using the formula

f = \phi/h

, where

\phi

is the work function and

h

is Planck's constant.

Solution approach

Correct Answer + Core Reason: $1.6 \times 10^{15} \mathrm{~Hz}$ . The threshold frequency is calculated using the work function and Planck's constant.

Step-by-Step Solution:

Step 1: Identify the given information: Work function $\phi = 6.68 \mathrm{eV}$ .

Step 2: Convert the work function from eV to Joules: $\phi = 6.68 \mathrm{eV} \times 1.602 \times 10^{-19} \mathrm{J/eV} = 1.069 \times 10^{-18} \mathrm{J}$ .

Step 3: Use the formula for threshold frequency: $f_0 = \frac{\phi}{h}$ , where $h = 6.626 \times 10^{-34} \mathrm{Js}$ is Planck's constant.

Step 4: Substitute the values: $f_0 = \frac{1.069 \times 10^{-18} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}$ .

Step 5: Calculate the threshold frequency: $f_0 = 1.613 \times 10^{15} \mathrm{Hz} \approx 1.6 \times 10^{15} \mathrm{Hz}$ .

Why other options are wrong:
- Option A: $8 \times 10^{15} \mathrm{~Hz}$ is incorrect, likely from a miscalculation or incorrect formula.
- Option C: $10 \times 10^{15} \mathrm{~Hz}$ is incorrect, likely from a miscalculation or incorrect formula.
- Option D: $4 \times 10^{15} \mathrm{~Hz}$ is incorrect, likely from a miscalculation or incorrect formula.

Key Concepts:
- Work function
- Threshold frequency
- Planck's constant
- Photoelectric effect

Memory Aids:
- $E = hf$ (Energy = Planck's constant * frequency)

Quick Tips:
- Always convert energy to Joules before using it in calculations.

28 Easy

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Application +4 | -1

The threshold frequency for a photosensitive metal is

3.3 \times {10^{14}}\,Hz.\,

If light of frequency

8.2 \times {10^{14}}\,Hz

is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

View answer and explanation

Correct option: A

2 V

Option A (2 V) is correct because the cut-off voltage (stopping potential) can be calculated using the photoelectric equation, and it's approximately 2V.

Solution approach

Correct Answer + Core Reason: 2 V. The cut-off voltage is calculated using the photoelectric equation.

Step-by-Step Solution:

Step 1: Recall the photoelectric equation: $eV_0 = h(f - f_0)$ , where $e$ is the elementary charge ( $1.6 \times 10^{-19} \text{{ C}}$ ), $V_0$ is the cut-off voltage, $h$ is Planck's constant ( $6.63 \times 10^{-34} \text{{ J s}}$ ), $f$ is the frequency of the incident light, and $f_0$ is the threshold frequency.

Step 2: Substitute the given values: $e V_0 = h (8.2 \times 10^{14} \text{{ Hz}} - 3.3 \times 10^{14} \text{{ Hz}})$ .

Step 3: Simplify: $e V_0 = h (4.9 \times 10^{14} \text{{ Hz}})$ .

Step 4: Calculate $V_0$ : $V_0 = \frac{{h (4.9 \times 10^{14} \text{{ Hz}})}}{e} = \frac{{6.63 \times 10^{-34} \text{{ J s}} \times 4.9 \times 10^{14} \text{{ Hz}}}}{{1.6 \times 10^{-19} \text{{ C}}}} \approx 2.03 \text{{ V}}$ .

Step 5: The cut-off voltage is approximately 2 V.

Why other options are wrong:
- Option B (3 V): This is incorrect based on the calculation.
- Option C (5 V): This is incorrect based on the calculation.
- Option D (1 V): This is incorrect based on the calculation.

Key Concepts:
- Photoelectric effect
- Cut-off voltage
- Photoelectric equation

Memory Aids:
- Remember the photoelectric equation: $eV_0 = h(f - f_0)$ .

Quick Tips:
- Use the photoelectric equation to calculate the cut-off voltage.

29 Medium

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Application +4 | -1

The electron in the hydrogen atom jumps from excited state

\left( {n = 3} \right)

to its ground state

\left( {n = 1} \right)

and the photons thus emitted irradiate a photosensitive material. If the work function of the material is

5.1\,\,\mathrm{eV}

, the stopping potential is estimated to be (the energy of the electron in

{n^{th}}

state

{E_n} = - \frac{{13.6}}{{{n^2}}}\,eV

)

View answer and explanation

Correct option: D

\begin{array}{llllllllllllllllllll} {\mathrm{7}\,\,\mathrm{V}} \end{array}

Option D (7 V) is correct because it is the stopping potential calculated using the energy difference between the excited and ground states and the work function.

Solution approach

Correct Answer + Core Reason: 7 V. The stopping potential is calculated by finding the energy of the emitted photon and subtracting the work function.

Step-by-Step Solution:

Step 1: Identify the given information: Electron jumps from $n = 3$ to $n = 1$ , work function $\phi = 5.1 \text{ eV}$ , and $E_n = -\frac{13.6}{n^2} \text{ eV}$ . We need to find the stopping potential $V_s$ .

Step 2: Calculate the energy of the electron in the $n = 3$ state: $E_3 = -\frac{13.6}{3^2} \text{ eV} = -\frac{13.6}{9} \text{ eV} \approx -1.51 \text{ eV}$ .

Step 3: Calculate the energy of the electron in the $n = 1$ state: $E_1 = -\frac{13.6}{1^2} \text{ eV} = -13.6 \text{ eV}$ .

Step 4: Calculate the energy of the emitted photon: $E_{\text{photon}} = E_3 - E_1 = -1.51 \text{ eV} - (-13.6 \text{ eV}) = 12.09 \text{ eV}$ .

Step 5: Apply the photoelectric effect equation: $E_{\text{photon}} = \phi + eV_s$ , where $V_s$ is the stopping potential.

Step 6: Substitute the values: $12.09 \text{ eV} = 5.1 \text{ eV} + eV_s$ .

Step 7: Solve for the stopping potential: $eV_s = 12.09 \text{ eV} - 5.1 \text{ eV} = 6.99 \text{ eV} \approx 7 \text{ eV}$ . Therefore, $V_s = 7 \text{ V}$ .

Step 8: Final Answer: 7 V

Why other options are wrong:
- Option A: $\mathrm{5}\mathrm{.1}\,\,\mathrm{V}$ : This is equal to the work function and doesn't account for the energy of the emitted photon.
- Option B: $\mathrm{12}\mathrm{.1}\,\,\mathrm{V}$ : This is approximately the energy of the emitted photon but doesn't account for the work function.
- Option C: $\mathrm{17}\mathrm{.2}\,\,\mathrm{V}$ : This value is likely obtained by adding the energy of the photon and the work function, which is incorrect.

Key Concepts:
- Energy levels in hydrogen atom
- Photoelectric effect
- Stopping potential
- Energy of a photon

Memory Aids:
- $E_n = -\frac{13.6}{n^2} \text{ eV}$
- $E_{\text{photon}} = E_f - E_i$
- $eV_s = E_{\text{photon}} - \phi$

Quick Tips:
- Remember the formula for energy levels in a hydrogen atom.
- Be careful with the signs when calculating energy differences.
- Ensure that the units are consistent throughout the calculation.

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Remembering +4 | -1

A stream of fast moving negatively charged particles are called

View answer and explanation

Correct option: A

Cathode rays

Option A (Cathode rays) is correct because cathode rays are streams of fast-moving, negatively charged particles (electrons).

Solution approach

Correct Answer + Core Reason: Cathode rays are streams of fast moving negatively charged particles.

Step-by-Step Solution:

Step 1: Recall the definition of cathode rays.

Step 2: Cathode rays are produced in vacuum tubes and are composed of electrons, which are negatively charged.

Step 3: Therefore, a stream of fast-moving negatively charged particles is called cathode rays.

Why other options are wrong:
- Option B: Hydrogen is a neutral atom or a positively charged ion (proton) when ionized.
- Option C: Helium is a neutral atom or a positively charged ion (alpha particle) when ionized.
- Option D: This is incorrect because cathode rays are a specific type of particle stream.

Key Concepts:
- Cathode rays
- Electrons
- Atomic structure

Memory Aids:
- Cathode rays are negatively charged, like electrons.

Quick Tips:
- Remember the historical significance of cathode rays in the discovery of the electron.

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