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

1 Easy

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.

2 Easy

2 of 22Question permalink

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.

3 Easy

3 of 22Question permalink

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.

4 Easy

4 of 22Question permalink

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.

5 Easy

5 of 22Question permalink

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.

6 Easy

6 of 22Question permalink

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.

7 Easy

7 of 22Question permalink

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.

8 Easy

8 of 22Question permalink

Understanding +4 | -1

The wavelength of the waves associated with a proton and a photon are the same. Therefore, the two have equal

View answer and explanation

Correct option: C

Momentum

Option C is correct because if the wavelengths are the same, then their momenta must be equal according to the de Broglie relation λ = h/p.

Solution approach

Correct Answer + Core Reason: Momentum

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength formula. The de Broglie wavelength (λ) is related to the momentum (p) by the equation: $\lambda = h/p$ , where h is Planck's constant.

Step 2: Apply the formula to both the proton and the photon. Let λ_p be the wavelength of the proton and λ_ph be the wavelength of the photon. Let p_p be the momentum of the proton and p_ph be the momentum of the photon. Then, λ_p = h/p_p and λ_ph = h/p_ph.

Step 3: Use the given information. We are given that λ_p = λ_ph.

Step 4: Equate the expressions for momentum. Since λ_p = λ_ph, we have h/p_p = h/p_ph. This implies that p_p = p_ph.

Why other options are wrong:
- Option A: Mass is not necessarily equal. Photons are massless.
- Option B: Velocity is not necessarily equal. Photons always travel at the speed of light.
- Option D: Kinetic energy is not necessarily equal. Kinetic energy depends on both mass and velocity, and these are not necessarily the same for the proton and photon.

Key Concepts:
- de Broglie wavelength
- Momentum

Memory Aids:
- If wavelengths are equal, momenta are equal (λ = h/p).

Quick Tips:
- Equal wavelengths imply equal momenta.

9 Easy

9 of 22Question permalink

Understanding +4 | -1

Which of the following statement is not true about stopping potential

\left(\mathrm{V}_0\right)?

View answer and explanation

Correct option: C

It increases with increase in intensity of the incident light

Option C is correct because stopping potential is independent of the intensity of incident light; it depends on the frequency and the emitter material.

Solution approach

Correct Answer + Core Reason: It increases with increase in intensity of the incident light. Stopping potential is independent of the intensity of light.

Step-by-Step Solution:

Step 1: Define stopping potential. Stopping potential is the minimum negative potential applied to the anode to stop the photoelectric current.

Step 2: Recall the factors affecting stopping potential. Stopping potential depends on the frequency of incident light and the work function of the emitter material, but not on the intensity of the incident light.

Step 3: Explain the effect of intensity. Increasing the intensity of light increases the number of photoelectrons emitted, but not their maximum kinetic energy. Since stopping potential is related to the maximum kinetic energy, it remains unchanged.

Step 4: Conclusion: Therefore, the statement that stopping potential increases with an increase in the intensity of incident light is not true.

Why other options are wrong:
- Option A: It depends on the nature of emitter material. This is true because the work function of the material affects the stopping potential.
- Option B: It depends upon frequency of the incident light. This is true because the kinetic energy of emitted electrons depends on the frequency of incident light.
- Option D: It is $1 / \mathrm{e}$ times the maximum kinetic energy of electrons emitted. This is related to the definition of stopping potential and is a true statement.

Key Concepts:
- Photoelectric effect
- Stopping potential
- Work function
- Intensity of light

Memory Aids:
- Stopping potential is like a \"brake\" that stops the electrons; intensity is like the \"number of cars,\" but the \"brake\" needed depends on the \"speed\" (frequency).

Quick Tips:
- Remember that intensity affects the number of photoelectrons, while frequency affects their kinetic energy and thus the stopping potential.

10 Easy

10 of 22Question permalink

Remembering +4 | -1

A photocell works on the principle of

View answer and explanation

Correct option: B

Photoelectric effect

Option B is correct because a photocell's operation is based on the photoelectric effect, where light incident on a material causes the emission of electrons.

Solution approach

Correct Answer + Core Reason: Photoelectric effect

Step-by-Step Solution:

Step 1: Define a photocell. A photocell is a device that converts light energy into electrical energy.

Step 2: Describe the photoelectric effect. The photoelectric effect is the emission of electrons when light shines on a material.

Step 3: Explain how a photocell works. In a photocell, light strikes a photosensitive material, causing electrons to be emitted. These electrons are then collected, creating an electric current.

Why other options are wrong:
- Option A: Bernoulli's principle deals with fluid dynamics, not the emission of electrons.
- Option C: Kirchhoff's rules are used for analyzing electrical circuits, not the emission of electrons.
- Option D: Radioactivity law describes the decay of radioactive materials, not the emission of electrons by light.

Key Concepts:
- Photoelectric effect
- Photocell

Memory Aids:
- Photo = light, electric = electricity. Photocell converts light to electricity via the photoelectric effect.

Quick Tips:
- The photoelectric effect is fundamental to the operation of photocells.

11 Easy

11 of 22Question permalink

Application +4 | -1

In photoelectric experiment energy of

2.48 \mathrm{~eV}

irradiates a photo sensitive material. The stopping potential was measured to be

0.5 \mathrm{~V}

. Work function of the photo sensitive material is :

View answer and explanation

Correct option: D

1.98 \mathrm{~eV}

Option D (

1.98 \mathrm{~eV}

) is correct because the work function can be calculated by subtracting the stopping potential energy from the incident photon energy.

Solution approach

Correct Answer + Core Reason: $1.98 \mathrm{~eV}$ . The work function is the difference between the incident energy and the stopping potential energy.

Step-by-Step Solution:

Step 1: Recall the photoelectric equation: $E = W_0 + eV_0$ , where $E$ is the energy of the incident photon, $W_0$ is the work function, $e$ is the elementary charge, and $V_0$ is the stopping potential.

Step 2: Substitute the given values: $2.48 \text{{ eV}} = W_0 + e(0.5 \text{{ V}})$ .

Step 3: Since $eV_0$ is in electron volts, we can directly substitute $eV_0 = 0.5 \text{{ eV}}$ .

Step 4: Solve for the work function: $W_0 = 2.48 \text{{ eV}} - 0.5 \text{{ eV}} = 1.98 \text{{ eV}}$ .

Why other options are wrong:
- **Option A ( $0.5 \mathrm{~eV}$ ): This is the stopping potential, not the work function.
- Option B ( $1.68 \mathrm{~eV}$ ): This is incorrect based on the calculation.
- Option C ( $2.48 \mathrm{~eV}$ ): This is the energy of the incident photon, not the work function.

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

Memory Aids:**
- Remember the photoelectric equation: $E = W_0 + eV_0$ .

Quick Tips:
- Use the photoelectric equation to calculate the work function.

12 Easy

12 of 22Question permalink

Application +4 | -1

de-Broglie wavelength of a proton and an electron is same. The ratio of kinetic energy of electron to that of proton is

View answer and explanation

Correct option: B

1835

Option B (1835) is correct because the ratio of kinetic energies is equal to the ratio of masses when de-Broglie wavelengths are the same.

Solution approach

Correct Answer + Core Reason: 1835. When the de-Broglie wavelengths are equal, the ratio of kinetic energies is the ratio of masses.

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: Since the de-Broglie wavelengths of the proton and electron are the same, we have $\lambda_e = \lambda_p$ .

Step 3: Therefore, $\frac{h}{\sqrt{2m_eK_e}} = \frac{h}{\sqrt{2m_pK_p}}$ .

Step 4: Squaring both sides and simplifying, we get $2m_eK_e = 2m_pK_p$ , which implies $\frac{K_e}{K_p} = \frac{m_p}{m_e}$ .

Step 5: The mass of a proton is approximately 1836 times the mass of an electron, so $\frac{K_e}{K_p} \approx 1836$ .

Step 6: Final Answer: The ratio of kinetic energy of the electron to that of the proton is approximately 1835.

Why other options are wrong:
- Option A (1): This would be true if the masses were equal, but they are not.
- **Option C ( $\frac{1}{1867}$ ): This is the inverse of the correct ratio, suggesting the calculation was inverted.
- Option D (933.5): This is approximately half the correct value, indicating a possible division by 2 error.

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

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

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

13 Easy

13 of 22Question permalink

Remembering +4 | -1

Kinetic mass of photon is

View answer and explanation

Correct option: B

h v / c^2

Option B (

h v / c^2

) is correct because it represents the kinetic mass of a photon derived from the energy-mass equivalence principle.

Solution approach

Correct Answer + Core Reason: $h v / c^2$ . The kinetic mass of a photon is derived from $E = mc^2$ and $E = hv$ .

Step-by-Step Solution:

Step 1: Recall the energy-mass equivalence relation: $E = mc^2$ , where $E$ is energy, $m$ is mass, and $c$ is the speed of light.

Step 2: Recall the energy of a photon: $E = hv$ , where $h$ is Planck's constant and $v$ is the frequency of the photon.

Step 3: Equate the two expressions for energy: $mc^2 = hv$ .

Step 4: Solve for the mass $m$ : $m = \frac{{hv}}{{c^2}}$ .

Step 5: The kinetic mass of the photon is therefore $m = \frac{{hv}}{{c^2}}$ .

Why other options are wrong:
- Option A (Zero): Photons have zero *rest* mass, but they have kinetic mass due to their energy.
- **Option C ( $h v^3 / c^2$ ): This is dimensionally incorrect and doesn't follow from the energy-mass equivalence.
- Option D ( $h v / {c}$ ): This has the wrong units and is not derived from the correct relationship between energy, mass, and frequency.

Key Concepts:**
- Energy-mass equivalence: $E = mc^2$
- Photon energy: $E = hv$

Memory Aids:
- Remember $E=mc^2$ and $E=hv$ to derive the kinetic mass of a photon.

Quick Tips:
- Always relate energy and mass using Einstein's equation when dealing with photons.

14 Easy

14 of 22Question permalink

Understanding +4 | -1

The wavelength of the matter waves is independent of

View answer and explanation

Correct option: A

Charge

Option A is correct because the de Broglie wavelength depends on momentum, velocity, and mass, but not on charge.

Solution approach

Correct Answer + Core Reason: Charge

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength formula. The de Broglie wavelength (λ) is given by: $\lambda = h/p$ , where h is Planck's constant and p is the momentum.

Step 2: Express momentum in terms of mass and velocity. Momentum (p) is given by: $p = mv$ , where m is the mass and v is the velocity.

Step 3: Substitute the expression for momentum into the de Broglie wavelength formula. Substituting $p = mv$ into $\lambda = h/p$ gives: $\lambda = h/(mv)$ .

Step 4: Analyze the formula. The formula $\lambda = h/(mv)$ shows that the de Broglie wavelength depends on mass (m) and velocity (v), but not on charge.

Why other options are wrong:
- Option B: The de Broglie wavelength depends on momentum.
- Option C: The de Broglie wavelength depends on velocity.
- Option D: The de Broglie wavelength depends on mass.

Key Concepts:
- de Broglie wavelength
- Momentum

Memory Aids:
- The de Broglie wavelength formula $\lambda = h/(mv)$ does not include charge.

Quick Tips:
- The charge of a particle does not affect its de Broglie wavelength.

15 Easy

15 of 22Question permalink

Application +4 | -1

The work function of a substance is

3.0 \mathrm{~eV}.

The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

View answer and explanation

Correct option: A

414 \mathrm{~nm}

Option A (

414 \mathrm{~nm}

) is correct because it is calculated using the formula relating work function and wavelength:

\lambda = \frac{hc}{W_0}

.

Solution approach

Correct Answer + Core Reason: $414 \mathrm{~nm}$ . The longest wavelength is calculated using the work function and the formula $\lambda = \frac{hc}{W_0}$ .

Step-by-Step Solution:

Step 1: Define the work function $W_0 = 3.0 \mathrm{~eV}$ .

Step 2: Convert the work function to Joules: $W_0 = 3.0 \mathrm{~eV} \times 1.602 \times 10^{-19} \mathrm{~J/eV} = 4.806 \times 10^{-19} \mathrm{~J}$ .

Step 3: Use the formula relating work function and wavelength: $W_0 = \frac{hc}{\lambda}$ , where $h = 6.626 \times 10^{-34} \mathrm{~J \cdot s}$ and $c = 3.0 \times 10^8 \mathrm{~m/s}$ .

Step 4: Rearrange the formula to solve for wavelength: $\lambda = \frac{hc}{W_0}$ .

Step 5: Substitute the values: $\lambda = \frac{(6.626 \times 10^{-34} \mathrm{~J \cdot s})(3.0 \times 10^8 \mathrm{~m/s})}{4.806 \times 10^{-19} \mathrm{~J}}$ .

Step 6: Calculate the wavelength: $\lambda = \frac{19.878 \times 10^{-26}}{4.806 \times 10^{-19}} \mathrm{~m} = 4.136 \times 10^{-7} \mathrm{~m}$ .

Step 7: Convert the wavelength to nanometers: $\lambda = 4.136 \times 10^{-7} \mathrm{~m} \times 10^9 \mathrm{~nm/m} = 413.6 \mathrm{~nm} \approx 414 \mathrm{~nm}$ .

Why other options are wrong:
- Option B (400 nm): This value is close, but not as accurate as the calculated value. It may result from rounding errors or using approximate values for constants.
- Option C (215 nm): This value is significantly lower than the correct wavelength, indicating a possible error in the calculation or misunderstanding of the relationship between work function and wavelength.
- Option D (200 nm): This value is also significantly lower than the correct wavelength, indicating a possible error in the calculation or misunderstanding of the relationship between work function and wavelength.

Key Concepts:
- Work function
- Photoelectric effect
- Relationship between energy and wavelength: $E = \frac{hc}{\lambda}$

Memory Aids:
- Remember the formula: $W_0 = \frac{hc}{\lambda}$

Quick Tips:
- Ensure consistent units (convert eV to Joules if necessary).
- Use accurate values for Planck's constant and the speed of light.

16 Easy

16 of 22Question permalink

Application +4 | -1

The threshold frequency of a metal with work function

6.63 \mathrm{~eV}

is:

View answer and explanation

Correct option: A

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

Option A (

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 $\phi = 6.63 \mathrm{eV}$ .

Step 2: Convert the work function from eV to Joules: $\phi = 6.63 \mathrm{eV} \times 1.602 \times 10^{-19} \mathrm{J/eV} = 1.062 \times 10^{-18} \mathrm{J}$ .

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: $f_0 = \frac{1.062 \times 10^{-18} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}$ .

Step 5: Calculate the threshold frequency: $f_0 = 1.603 \times 10^{15} \mathrm{Hz} \approx 1.6 \times 10^{15} \mathrm{Hz}$ .

Why other options are wrong:
- Option B: $16 \times 10^{12} \mathrm{~Hz}$ is incorrect, likely from a miscalculation or incorrect formula.
- Option C: $1.6 \times 10^{12} \mathrm{~Hz}$ is incorrect, likely from a miscalculation or incorrect formula.
- Option D: $16 \times 10^{15} \mathrm{~Hz}$ 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.

17 Easy

17 of 22Question permalink

Application +4 | -1

An

\alpha

particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

View answer and explanation

Correct option: B

2 \sqrt{2}: 1

Option B (

2 \sqrt{2}: 1

) is correct because the momentum is proportional to the square root of the kinetic energy, which is proportional to the charge and potential difference.

Solution approach

Correct Answer + Core Reason: $2 \sqrt{2}: 1$ . The ratio of linear momenta is determined by the charge and mass of the particles and the potential difference.

Step-by-Step Solution:

Step 1: Recall the relationship between kinetic energy (KE) and momentum (p): $KE = p^2 / (2m)$ , so $p = \sqrt{2mKE}$ .

Step 2: The kinetic energy gained by a charged particle accelerated through a potential difference V is $KE = qV$ .

Step 3: Therefore, $p = \sqrt{2mqV}$ .

Step 4: For the alpha particle, $q_\alpha = 2e$ and $m_\alpha = 4m_p$ , where $m_p$ is the mass of the proton.

Step 5: For the proton, $q_p = e$ and $m_p = m_p$ .

Step 6: The ratio of momenta is: $\frac{p_\alpha}{p_p} = \frac{\sqrt{2m_\alpha q_\alpha V}}{\sqrt{2m_p q_p V}} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{(4m_p)(2e)}{m_p e}} = \sqrt{8} = 2\sqrt{2}$ .

Step 7: The ratio is $2\sqrt{2}: 1$ .

Why other options are wrong:
- Option A: $\sqrt{2}: 1$ is incorrect; it doesn't account for both the mass and charge differences.
- Option C: $4\sqrt{2}: 1$ is incorrect; the calculation is flawed.
- Option D: $8: 1$ is incorrect; the square root was not taken correctly.

Key Concepts:
- Kinetic energy
- Momentum
- Potential difference
- Alpha particle

Memory Aids:
- $KE = qV$ and $p = \sqrt{2mKE}$

Quick Tips:
- Remember to consider both mass and charge when comparing particles accelerated through the same potential.

18 Easy

18 of 22Question permalink

Application +4 | -1

When UV light of wavelength

300 \mathrm{~nm}

is incident on the metal surface having work function

2.13 \mathrm{~eV}

, electron emission takes place. The stopping potential is :

(

Given

\mathrm{hc}=1240 \mathrm{~eV} \mathrm{~nm})

View answer and explanation

Correct option: C

2 V

Option C (2 V) is correct because the stopping potential is calculated using the photoelectric effect equation.

Solution approach

Correct Answer + Core Reason: 2 V. The stopping potential is calculated using the photoelectric effect equation.

Step-by-Step Solution:

Step 1: Calculate the energy of the incident photon: $E = hc/\lambda = 1240 \text{{ eV nm}} / 300 \text{{ nm}} \approx 4.13 \text{{ eV}}$ .

Step 2: Use the photoelectric equation: $eV_s = E - \phi$ , where $V_s$ is the stopping potential, $E$ is the photon energy, and $\phi$ is the work function.

Step 3: Substitute the given values: $eV_s = 4.13 \text{{ eV}} - 2.13 \text{{ eV}} = 2 \text{{ eV}}$ .

Step 4: Solve for the stopping potential: $V_s = 2 \text{{ eV}} / e = 2 \text{{ V}}$ .

Why other options are wrong:
- Option A: 4 V is incorrect due to calculation errors.
- Option B: 4.1 V is incorrect due to calculation errors.
- Option D: 1.5 V is incorrect due to calculation errors.

Key Concepts:
- Photoelectric effect
- Stopping potential
- Work function
- Photon energy

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

Quick Tips:
- Remember to use consistent units (eV).

19 Easy

19 of 22Question permalink

Application +4 | -1

In a photoelectric effect experiment a light of
frequency 1.5 times the threshold frequency is
made to fall on the surface of photosensitive
material. Now if the frequency is halved and
intensity is doubled, the number of photo electrons
emitted will be

View answer and explanation

Correct option: D

zero

Option D (zero) is correct because if the frequency is halved, it becomes less than the threshold frequency, and no photoelectrons are emitted.

Solution approach

Correct Answer + Core Reason: zero. If the frequency is halved and becomes less than the threshold frequency, no photoelectrons will be emitted.

Step-by-Step Solution:

Step 1: Let the threshold frequency be $f_0$ .

Step 2: The initial frequency is $f_1 = 1.5f_0$ , which is greater than the threshold frequency, so photoelectrons are emitted.

Step 3: The new frequency is $f_2 = f_1/2 = (1.5f_0)/2 = 0.75f_0$ , which is less than the threshold frequency.

Step 4: Since the frequency is less than the threshold frequency, no photoelectrons will be emitted, regardless of the intensity.

Why other options are wrong:
- Option A: halved is incorrect because no photoelectrons are emitted.
- Option B: doubled is incorrect because no photoelectrons are emitted.
- Option C: quadrupled is incorrect because no photoelectrons are emitted.

Key Concepts:
- Photoelectric effect
- Threshold frequency
- Intensity

Memory Aids:
- Photoelectric emission only occurs if the frequency is greater than the threshold frequency.

Quick Tips:
- Always check if the frequency is above the threshold frequency before considering intensity.

20 Easy

20 of 22Question permalink

Understanding +4 | -1

Which of the following phenomena does not explain by wave nature of light.

(A) reflection

\quad

(B) diffraction

(C) photoelectric effect

\quad

(D) interference

\quad

(E) polarization

Choose the most appropriate answer from the options given below :

View answer and explanation

Correct option: B

C only

Option B (C only) is correct because the photoelectric effect is explained by the particle nature of light, while reflection, diffraction, interference, and polarization are explained by the wave nature of light.

Solution approach

Correct Answer + Core Reason: C only. Photoelectric effect is explained by the particle nature of light.

Step-by-Step Solution:

Step 1: Recall the wave nature of light. Reflection, diffraction, interference, and polarization are phenomena that can be explained by the wave nature of light.

Step 2: Recall the particle nature of light. The photoelectric effect is a phenomenon that can only be explained by the particle nature of light, where light is considered as a stream of photons.

Step 3: Identify the exception. The question asks for the phenomenon not explained by the wave nature of light. Therefore, the photoelectric effect is the correct answer.

Why other options are wrong:
- Option A: E only. Polarization is explained by the wave nature of light.
- Option C: B, D only. Both diffraction and interference are explained by the wave nature of light.
- Option D: A, C only. Reflection is explained by the wave nature of light, while the photoelectric effect is not.

Key Concepts:
- Wave nature of light
- Particle nature of light
- Reflection
- Diffraction
- Interference
- Polarization
- Photoelectric effect

Memory Aids:
- Wave phenomena: think of \"WavIng DIffuse Reflections and Polarization\" (WIDRP)
- Photoelectric effect: think of \"Particles Ejecting Electrons\" (PEE)

Quick Tips:
- Remember that the photoelectric effect is a key piece of evidence for the particle nature of light.

21 Easy

21 of 22Question permalink

Application +4 | -1

A particle at rest disintegrates into two fragments of masses in the ratio of 1: 2 having non-zero velocities. The ratio of the de-Broglie wavelength of the lighter particle to that of the heavier particle is

View answer and explanation

Correct option: A

1

Option A (1) is correct because, due to conservation of momentum, the magnitudes of the momenta of the two fragments are equal, resulting in equal de Broglie wavelengths.

Solution approach

Correct Answer + Core Reason: 1. The de-Broglie wavelengths are equal due to conservation of momentum.

Step-by-Step Solution:

Step 1: State the conservation of momentum: Since the particle was initially at rest, the total momentum of the fragments must be zero.

Step 2: Let $p_1$ and $p_2$ be the momenta of the lighter and heavier particles, respectively. Then $p_1 + p_2 = 0$ , which means $|p_1| = |p_2|$ .

Step 3: Recall the de-Broglie wavelength formula: $\lambda = \frac{h}{p}$ , where $h$ is Planck's constant and $p$ is momentum.

Step 4: Let $\lambda_1$ and $\lambda_2$ be the de-Broglie wavelengths of the lighter and heavier particles, respectively. Then $\lambda_1 = \frac{h}{|p_1|}$ and $\lambda_2 = \frac{h}{|p_2|}$ .

Step 5: Find the ratio of the wavelengths: $\frac{\lambda_1}{\lambda_2} = \frac{h/|p_1|}{h/|p_2|} = \frac{|p_2|}{|p_1|}$ .

Step 6: Since $|p_1| = |p_2|$ , the ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = 1$ .

Why other options are wrong:
- Option B (0): This would imply that one of the particles has infinite momentum, which is not possible.
- Option C (0): Same as Option B.
- Option D (0): Same as Option B.

Key Concepts:
- Conservation of momentum
- de-Broglie wavelength: $\lambda = \frac{h}{p}$

Memory Aids:
- Remember that in a closed system, momentum is conserved.

Quick Tips:
- When a particle at rest decays into two fragments, the magnitudes of their momenta are equal.

22 Easy

22 of 22Question permalink

Understanding +4 | -1

Statement-1: Photoelectric effect establishes quantum nature of light.
Statement-2: There is negligible time lag between photon collisions with the material and photoelectron emission irrespective of intensity of incident light. (Assume incident light is of frequency greater than threshold frequency of the material).

View answer and explanation

Correct option: A

Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1

Option A (Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1) is correct because the instantaneous emission of photoelectrons, regardless of intensity, demonstrates the particle nature of light.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1. The instantaneous emission of photoelectrons supports the quantum nature of light.

Step-by-Step Solution:

Step 1: Analyze Statement-1: Photoelectric effect establishes quantum nature of light. This is true because the photoelectric effect can only be explained by considering light as consisting of discrete packets of energy (photons), rather than as a continuous wave.

Step 2: Analyze Statement-2: There is negligible time lag between photon collisions with the material and photoelectron emission irrespective of intensity of incident light. This is also true. The emission of photoelectrons is almost instantaneous, even at low intensities, which is a key piece of evidence for the quantum nature of light.

Step 3: Determine the relationship between Statement-1 and Statement-2: The negligible time lag (Statement-2) is a direct consequence of the quantum nature of light (Statement-1). If light were a continuous wave, we would expect a time lag for the electron to absorb enough energy to be emitted. The fact that emission is instantaneous, regardless of intensity, supports the idea that light comes in discrete packets (photons) that can transfer their energy to electrons in a single interaction.

Final Answer: Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1

Why other options are wrong:
- Option B (Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1): This is incorrect because Statement-2 directly explains why Statement-1 is true.
- Option C (Statement-1 is true, Statement-2 is false): This is incorrect because Statement-2 is true.
- Option D (Statement-1 is false, Statement-2 is true): This is incorrect because Statement-1 is true.

Key Concepts:
- Photoelectric effect
- Quantum nature of light
- Photons

Memory Aids:
- Remember that the instantaneous emission of photoelectrons is a key piece of evidence for the quantum nature of light.

Quick Tips:
- Understand the key experimental observations that support the quantum nature of light.

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