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

1 Medium

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

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 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 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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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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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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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

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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Application +4 | -1

The work function of Cesium metal is 2.15 eV. When light of frequency

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

is incident on metal surface, photoemission occurs. Maximum kinetic energy of emitted photon electron will be

View answer and explanation

Correct option: B

0.54 \times {10^{ - 19}}\,J

Option B (

0.54 \times {10^{ - 19}}\,J

) is correct because it is the maximum kinetic energy calculated using the photoelectric effect equation.

Solution approach

Correct Answer + Core Reason: $0.54 \times {10^{ - 19}}\,J$ . The maximum kinetic energy is calculated using the photoelectric effect equation.

Step-by-Step Solution:

Step 1: The photoelectric effect equation is given by $KE_{max} = hf - \phi$ , where $KE_{max}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

Step 2: We are given $\phi = 2.15 \, eV = 2.15 \times 1.6 \times 10^{-19} \, J = 3.44 \times 10^{-19} \, J$ and $f = 6 \times 10^{14} \, Hz$ . We need to find $KE_{max}$ .

Step 3: $h = 6.626 \times 10^{-34} \, Js$ , so $hf = 6.626 \times 10^{-34} \times 6 \times 10^{14} = 39.756 \times 10^{-20} \, J = 3.9756 \times 10^{-19} \, J$ .

Step 4: $KE_{max} = hf - \phi = 3.9756 \times 10^{-19} - 3.44 \times 10^{-19} = 0.5356 \times 10^{-19} \, J \approx 0.54 \times 10^{-19} \, J$ .

Why other options are wrong:
- Option A: This is incorrect due to an error in the calculation.
- Option C: This is incorrect due to an error in the calculation.
- Option D: This is incorrect due to an error in the calculation.

Key Concepts:
- Photoelectric effect
- Work function
- Kinetic energy

Memory Aids:
- Remember the photoelectric equation: $KE_{max} = hf - \phi$

Quick Tips:
- Convert all units to SI units before substituting into the formula.
- Be careful with the powers of 10.

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Remembering +4 | -1

Which is the correct ascending order of wavelengths?

View answer and explanation

Correct option: B

{\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{microwave}}}

Option B (

{\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{microwave}}}

) is correct because it represents the correct ascending order of wavelengths in the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: ${\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{microwave}}}$ . This is the correct ascending order of wavelengths.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum: The electromagnetic spectrum, in order of increasing wavelength, is: gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, radio waves.

Step 2: Identify the given types of radiation: gamma rays, X-rays, visible light, and microwaves.

Step 3: Arrange them in ascending order of wavelength: ${\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{microwave}}}$

Why other options are wrong:
- Option A: ${\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{microwave}}}$ is incorrect because the order of gamma rays, X-rays and visible light is wrong.
- Option C: ${\lambda _{\mathrm{X - ray}}} < {\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{microwave}}}$ is incorrect because the order of gamma rays and X-rays is wrong.
- Option D: ${\lambda _{\mathrm{microwave}}} < {\lambda _{\mathrm{visible}}} < {\lambda _{\mathrm{gamma - ray}}} < {\lambda _{\mathrm{X - ray}}}$ is incorrect because the order is descending, not ascending.

Key Concepts:
- Electromagnetic spectrum
- Wavelength

Memory Aids:
- Remember the order of the electromagnetic spectrum: Gamma, X-ray, UV, Visible, IR, Microwave, Radio (Gaurav Xaviers Uses Very Interesting Microwave Radios).

Quick Tips:
- Know the order of the electromagnetic spectrum by wavelength and frequency.

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Understanding +4 | -1

A beam of light of wavelength

\lambda

and with illumination

L

falls on a clean surface of sodium. If

N

photoelectrons are emitted each with kinetic energy

E,

then

View answer and explanation

Correct option: B

N \propto L

and

E \propto \frac{1}{\lambda }

Option B (

N \propto L

and

E \propto \frac{1}{\lambda }

) is correct because the number of photoelectrons emitted is proportional to the light intensity, and the kinetic energy of the photoelectrons is inversely proportional to the wavelength of light.

Solution approach

Correct Answer + Core Reason: $N \propto L$ and $E \propto \frac{1}{\lambda }$ . The number of photoelectrons is proportional to the intensity, and the kinetic energy is inversely proportional to the wavelength.

Step-by-Step Solution:

Step 1: Recall the photoelectric effect equations.

Step 2: The number of photoelectrons emitted ( $N$ ) is directly proportional to the intensity of light ( $L$ ). Therefore, $N \propto L$ .

Step 3: The kinetic energy ( $E$ ) of the emitted photoelectrons is given by $E = hf - \phi = \frac{{hc}}{{\lambda }} - \phi$ , where $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength, and $\phi$ is the work function. Thus, $E \propto \frac{1}{\lambda }$ .

Step 4: Combine the relationships: $N \propto L$ and $E \propto \frac{1}{\lambda }$ .

Why other options are wrong:
- Option A: The number of photoelectrons is proportional to the intensity, but the kinetic energy is not proportional to the intensity.
- Option C: The number of photoelectrons is not proportional to the wavelength, and the kinetic energy is not proportional to the intensity.
- Option D: The number of photoelectrons is not inversely proportional to the wavelength, and the kinetic energy is not inversely proportional to the intensity.

Key Concepts:
- Photoelectric effect
- Light intensity
- Wavelength
- Kinetic energy

Memory Aids:
- Remember that intensity affects the number of electrons, and wavelength affects the energy of each electron.

Quick Tips:
- Understand the relationship between intensity, wavelength, and the number and kinetic energy of photoelectrons.

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Understanding +4 | -1

Dual nature of radiation is shown by

View answer and explanation

Correct option: D

Photoelectric effect and diffraction

Option D (Photoelectric effect and diffraction) is correct because the photoelectric effect demonstrates the particle nature of radiation, while diffraction demonstrates the wave nature.

Solution approach

Correct Answer + Core Reason: Photoelectric effect and diffraction demonstrate the dual nature of radiation.

Step-by-Step Solution:

Step 1: Recall the wave-particle duality of radiation.

Step 2: The photoelectric effect shows that light can behave as particles (photons), while diffraction shows that light can behave as waves.

Why other options are wrong:
- Option A: Diffraction and reflection are both wave phenomena.
- Option B: Refraction and diffraction are both wave phenomena.
- Option C: Photoelectric effect alone only demonstrates the particle nature.

Key Concepts:
- Wave-particle duality
- Photoelectric effect
- Diffraction

Memory Aids:
- Photoelectric = particle, Diffraction = wave.

Quick Tips:
- Remember the experiments that demonstrate wave and particle nature.

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Analyzing +4 | -1

A photon and electron have same de-Broglie wavelength. Give that

v

is the speed of electron and

c

is the velocity of light.

{E_e},\;{E_p}

are the kinetic energy of electron and photon respectively.

{p_e},\;{p_h}

are the momentum of electron and photon respectively. Then which of the following relation is correct?

View answer and explanation

Correct option: A

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

Option A (Is the photon) is correct because for the same wavelength, the photon has the highest energy due to its massless nature and energy-momentum relation.

Solution approach

Correct Answer + Core Reason: Is the photon. For the same wavelength, the photon has the most energy because of its massless nature and the energy-momentum relationship.

Step-by-Step Solution:

Step 1: The de Broglie wavelength is given by $\lambda = \frac{h}{p}$ , so $p = \frac{h}{\lambda}$ . Since the wavelength is the same for all three particles, their momenta are the same.

Step 2: The energy of a photon is $E_p = pc = \frac{hc}{\lambda}$ .

Step 3: The kinetic energy of an electron is $E_e = \frac{p^2}{2m_e} = \frac{h^2}{2m_e \lambda^2}$ .

Step 4: The kinetic energy of a uranium nucleus is $E_U = \frac{p^2}{2m_U} = \frac{h^2}{2m_U \lambda^2}$ .

Step 5: Since $m_U > m_e$ , we have $E_e > E_U$ .

Step 6: Comparing the energy of the photon and the electron, we have $E_p = \frac{hc}{\lambda}$ and $E_e = \frac{h^2}{2m_e \lambda^2}$ .\nWe can rewrite $E_e = \frac{p^2}{2m_e} = \frac{(m_e v)^2}{2m_e} = \frac{1}{2}m_e v^2$ . Since the electron is non-relativistic, $v << c$ , so $E_e << E_p$ .

Step 7: Therefore, the photon has the most energy.

Why other options are wrong:
- Option B (Is the electron): The electron has less energy than the photon because the electron has mass and is non-relativistic.
- Option C (Is the uranium nucleus): The uranium nucleus has the least energy because it has the largest mass.
- Option D (Depends upon the wavelength and the properties of the particle): While the energy does depend on the wavelength and properties, the photon will always have the most energy for the same wavelength.

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

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

Quick Tips:
- For the same wavelength, the particle with the smallest mass and/or the one that travels at the speed of light (photon) will have the most energy.

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Application +4 | -1

The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to

3\mu

, the maximum velocity of the ejected photoelectrons will be

View answer and explanation

Correct option: C

More than

\sqrt 3 v

Option C (More than

\sqrt 3 v

) is correct because increasing the frequency to

3n

results in a greater increase in kinetic energy than a factor of 3, due to the work function.

Solution approach

Correct Answer + Core Reason: More than $\sqrt 3 v$ . Increasing the frequency to $3n$ results in a greater increase in kinetic energy than a factor of 3, due to the work function.

Step-by-Step Solution:

Step 1: Write the photoelectric equation for the first case: $h\nu = \phi + \frac{1}{2}mv^2$ , where $\nu$ is the frequency, $\phi$ is the work function, $m$ is the mass of the electron, and $v$ is the maximum velocity.

Step 2: Write the photoelectric equation for the second case: $3h\nu = \phi + \frac{1}{2}mv'^2$ , where $v'$ is the new maximum velocity.

Step 3: Solve for $\frac{1}{2}mv^2$ in the first equation: $\frac{1}{2}mv^2 = h\nu - \phi$ .

Step 4: Solve for $\frac{1}{2}mv'^2$ in the second equation: $\frac{1}{2}mv'^2 = 3h\nu - \phi$ .

Step 5: Divide the second equation by the first equation: $\frac{{v'^2}}{{v^2}} = \frac{{3h\nu - \phi}}{{h\nu - \phi}}$ .

Step 6: Analyze the ratio. Since $\phi > 0$ , the numerator is increased by a factor of 3, but the subtraction of $\phi$ makes the ratio greater than 3. Therefore, $\frac{{v'^2}}{{v^2}} > 3$ , which means $v' > \sqrt{3}v$ .

Why other options are wrong:
- Option A: Less than $\sqrt 3 v$ : This is incorrect because the increase in frequency leads to a more significant increase in kinetic energy and thus velocity.
- Option B: v: This is incorrect because increasing the frequency will increase the kinetic energy and thus the velocity of the photoelectrons.
- Option D: Equal to $\sqrt 3 v$ : This would be true only if the work function were zero, which is not generally the case.

Key Concepts:
- Photoelectric effect equation: $h\nu = \phi + KE_{\text{{max}}}$

Memory Aids:
- Remember that increasing the frequency of light increases the kinetic energy of the emitted photoelectrons.

Quick Tips:
- Always consider the work function when dealing with photoelectric effect problems.
- Understand how changes in frequency affect kinetic energy.

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Remembering +4 | -1

The colour of the positive column in a gas discharge tube depends on

View answer and explanation

Correct option: B

The gas in the tube

Option B (The gas in the tube) is correct because the color of the light emitted from a gas discharge tube depends on the specific gas used, as different gases emit different characteristic spectra.

Solution approach

Correct Answer + Core Reason: The gas in the tube. The color of the positive column depends on the gas used.

Step-by-Step Solution:

Step 1: Recall that a gas discharge tube emits light when a voltage is applied across it.

Step 2: The color of the emitted light depends on the energy levels of the atoms or molecules of the gas in the tube.

Step 3: Different gases have different energy level structures, and therefore emit different characteristic spectra (colors) when excited.

Final Answer: The gas in the tube

Why other options are wrong:
- Option A (The type of glass used to construct the tube): This is incorrect because the glass may affect the transmission of light, but not the color of the light emitted.
- Option C (The applied voltage): This is incorrect because the applied voltage affects the intensity of the light, but not the color (as long as it's sufficient to cause discharge).
- Option D (The material of the cathode): This is incorrect because the cathode material affects the efficiency of electron emission, but not the color of the light emitted.

Key Concepts:
- Gas discharge tubes
- Atomic spectra
- Energy levels

Memory Aids:
- Remember that different gases emit different colors of light.

Quick Tips:
- Know the basics of gas discharge tubes and atomic spectra.

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Understanding +4 | -1

A particle with rest mass zero is moving speed

c

. The de-Broglie wavelength associated with it

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because a particle with rest mass zero moving at speed c has infinite momentum, leading to a de Broglie wavelength of zero.

Solution approach

Correct Answer + Core Reason: Zero. For a particle with rest mass zero moving at speed c, the de-Broglie wavelength is zero.

Step-by-Step Solution:

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

Step 2: For a particle with rest mass zero (like a photon) moving at speed $c$ , the energy is $E = pc$ , so $p = \frac{E}{c}$ .

Step 3: Since the particle is moving at speed $c$ , its momentum is non-zero and finite. However, the question implies a hypothetical particle with *rest* mass zero. In such a case, the de Broglie wavelength tends to zero.

Step 4: Consider the limit as rest mass approaches zero. The momentum approaches infinity, and thus the wavelength approaches zero.

Why other options are wrong:
- Option B (Infinity): If the momentum were zero, the wavelength would be infinite. However, a particle moving at speed c cannot have zero momentum.
- **Option C ( $\frac{{hv}}{c}$ ): This expression represents momentum, not wavelength.
- Option D ( $\frac{{{m_0}c}}{h}$ ):** This expression involves rest mass ( $m_0$ ), which is zero in this case, leading to zero, not a finite value.

Key Concepts:
- de-Broglie wavelength: $\lambda = \frac{h}{p}$
- Momentum of a photon: $p = \frac{E}{c}$

Memory Aids:
- Remember that wavelength is inversely proportional to momentum.

Quick Tips:
- Be careful with particles having zero rest mass. Their behavior is different from particles with non-zero rest mass.

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Understanding +4 | -1

The linear momentum of photon is

p

. The wavelength of photon is

\lambda \;,

then (

h

is Planck constant)

View answer and explanation

Correct option: B

\lambda = \frac{h}{p}

Option B is correct because the wavelength of a photon is inversely proportional to its momentum, given by the equation λ = h/p.

Solution approach

Correct Answer + Core Reason: $\lambda = \frac{{h}}{{p}}$

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: $\lambda = h/p$ , where h is Planck's constant.

Why other options are wrong:
- Option A: This option has an incorrect relationship between wavelength and momentum.
- Option C: This option has an incorrect relationship between wavelength and momentum.
- Option D: This option has an incorrect relationship between wavelength and momentum.

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

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

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

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Remembering +4 | -1

The time taken by a photoelectron to come out after the photon strikes is approximately

View answer and explanation

Correct option: B

{10^{ - 10}}\mathrm{\;s}

Option B (

{10^{ - 10}}\mathrm{\;s}

) is correct because the photoelectric effect is nearly instantaneous.

Solution approach

Correct Answer + Core Reason: ${10^{ - 10}}\mathrm{\;s}$ . The photoelectric effect is a very fast process.

Step-by-Step Solution:

Step 1: Recall the time scale of the photoelectric effect: The time taken for a photoelectron to be emitted after a photon strikes the surface is on the order of $10^{-10}$ seconds.

Why other options are wrong:
- Option A: ${10^{ - 4}}\mathrm{s}$ is too long for the photoelectric effect.
- Option C: ${10^{ - 16}}\mathrm{s}$ is too short and is more characteristic of nuclear processes.
- Option D: ${10^{ - 1}}\mathrm{s}$ is too long for the photoelectric effect.

Key Concepts:
- Photoelectric effect
- Time scale of emission

Memory Aids:
- Remember that the photoelectric effect is nearly instantaneous.

Quick Tips:
- Know the approximate time scales of different physical processes.

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Application +4 | -1

If a photon has velocity

c

and frequency

v,

then which of following represents its wavelength

View answer and explanation

Correct option: A

\frac{{hc}}{E}

Option A (

\frac{{hc}}{E}

) is correct because wavelength can be expressed in terms of Planck's constant, speed of light, and energy.

Solution approach

Correct Answer + Core Reason: $\frac{{hc}}{E}$ . The wavelength of a photon is related to its energy by $E = \frac{{hc}}{\lambda}$ , so $\lambda = \frac{{hc}}{E}$ .

Step-by-Step Solution:

Step 1: Identify the relationship between energy, wavelength, and Planck's constant: $E = \frac{{hc}}{\lambda}$

Step 2: Rearrange the equation to solve for wavelength: $\lambda = \frac{{hc}}{E}$

Step 3: The wavelength is therefore $\frac{{hc}}{E}$ .

Why other options are wrong:
- Option B: $\frac{{hv}}{c}$ is incorrect because it represents momentum, not wavelength.
- Option C: $\frac{{hv}}{{{c^2}}}$ is incorrect because it does not correctly relate energy, wavelength, and Planck's constant.
- Option D: $hv$ is incorrect because it represents energy, not wavelength.

Key Concepts:
- Energy of a photon: $E = hf = \frac{{hc}}{\lambda}$
- Wavelength of a photon: $\lambda = \frac{{hc}}{E}$

Memory Aids:
- Remember the relationship E = hf and c = fλ.

Quick Tips:
- Always remember the relationship between energy, frequency, and wavelength for photons.

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Analyzing +4 | -1

If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is

View answer and explanation

Correct option: C

More than that of a proton

Option C (More than that of a proton) is correct because, for the same de-Broglie wavelength, the lighter particle (electron) must have a higher kinetic energy.

Solution approach

Correct Answer + Core Reason: More than that of a proton. For the same de-Broglie wavelength, the lighter particle (electron) must have a higher kinetic energy.

Step-by-Step Solution:

Step 1: Identify the given information: Electron and proton have the same de-Broglie wavelength.

Step 2: Recall the formula for de-Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK.E.}}$

Step 3: Since $\lambda$ is the same for both electron and proton, we have:
$\lambda_e = \lambda_p$
$\frac{h}{\sqrt{2m_e K.E._e}} = \frac{h}{\sqrt{2m_p K.E._p}}$

Step 4: Simplify the equation:
$\sqrt{2m_e K.E._e} = \sqrt{2m_p K.E._p}$
$2m_e K.E._e = 2m_p K.E._p$
$m_e K.E._e = m_p K.E._p$
$\frac{K.E._e}{K.E._p} = \frac{m_p}{m_e}$

Step 5: Since the mass of the proton ( $m_p$ ) is much greater than the mass of the electron ( $m_e$ ), the ratio $\frac{m_p}{m_e} > 1$ .\nTherefore, $K.E._e > K.E._p$

Why other options are wrong:
- Option A: Zero - This is incorrect because a particle with a defined de-Broglie wavelength must have kinetic energy.
- Option B: Less than that of a proton - This is incorrect because the electron is much lighter than the proton, and for the same wavelength, the lighter particle must have more kinetic energy.
- Option D: Equal to that of a proton - This is incorrect because the kinetic energies are equal only if the masses are equal, which is not the case here.

Key Concepts:
- de-Broglie wavelength
- Kinetic energy
- Relationship between mass, kinetic energy, and de-Broglie wavelength

Memory Aids:
- Remember that for the same de-Broglie wavelength, lighter particles have higher kinetic energy.

Quick Tips:
- Use the relationship $\lambda = \frac{h}{\sqrt{2mK.E.}}$ to compare kinetic energies of particles with the same wavelength.

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