In a photoelectric experiment, the frequency of light is kept fixed above threshold while its intensity is doubled. What happens to the saturation current and the stopping potential?

Saturation current doubles; stopping potential is unchanged

Both saturation current and stopping potential double

Both saturation current and stopping potential are unchanged

Saturation current is unchanged; stopping potential doubles

Light below a metal's threshold frequency falls on it, and its intensity is increased many times. What happens to photoelectric emission?

No emission occurs at any intensity

Emission begins after a long delay

Emission occurs with zero kinetic energy

Emission begins once the intensity is high enough

Dual Nature of Radiation & Matter

10 NEET-UG practice questions on Dual Nature of Radiation & Matter, part of Physics. The first 10 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Medium
A deuteron and an α-particle, both initially at rest, are accelerated through the same potential difference. What is the ratio of the de Broglie wavelength of the deuteron to that of the α-particle?
1. A4
2. B2
3. C1/2
4. D√2
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Answer: B · 2
λ = h/√(2mqV), so at the same V, λ is proportional to 1/√(mq). Deuteron: mq = 2 × 1 = 2. α-particle: mq = 4 × 2 = 8. The ratio is √(8/2) = 2.
- A: The square root was not taken.
- C: The ratio has been inverted.
- D: The factor from the charge was left out, using 1/√m only.
Remember this
Through the same potential, de Broglie wavelength varies as 1/√(mq), so a larger mass and charge both shorten it.
Question 2 · Hard
A metal is lit in turn by light of wavelength 200 nm and of 400 nm. The stopping potential with 200 nm is three times that with 400 nm. Taking hc = 1240 eV nm, find the work function of the metal.
1. A4.65 eV
2. B3.1 eV
3. C7.75 eV
4. D1.55 eV
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Answer: D · 1.55 eV
Photon energies are 1240/200 = 6.2 eV and 1240/400 = 3.1 eV. Then 6.2 - φ = 3(3.1 - φ), so 2φ = 3.1 and φ = 1.55 eV. Check: stopping potentials 4.65 V and 1.55 V, ratio 3.
- A: This is the stopping energy with 200 nm light.
- B: This is the photon energy of 400 nm light, not the work function.
- C: The factor of 3 was applied to the wrong wavelength.
Remember this
Write eV₀ = hc/λ - φ for each wavelength and solve the pair; the work function does not change.
Question 3 · Medium
In a photoelectric experiment, the frequency of light is kept fixed above threshold while its intensity is doubled. What happens to the saturation current and the stopping potential?
1. ABoth saturation current and stopping potential double
2. BSaturation current doubles; stopping potential is unchanged
3. CBoth saturation current and stopping potential are unchanged
4. DSaturation current is unchanged; stopping potential doubles
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Answer: B · Saturation current doubles; stopping potential is unchanged
Doubling intensity doubles the photons per second, and so the electrons emitted per second and the saturation current. The stopping potential depends on hν - φ, which has not changed.
- A: Stopping potential does not depend on intensity.
- C: More photons per second do raise the saturation current.
- D: This reverses the roles of intensity and frequency.
Remember this
Intensity controls photocurrent; frequency and work function control stopping potential.
Question 4 · Easy
Light below a metal's threshold frequency falls on it, and its intensity is increased many times. What happens to photoelectric emission?
1. AEmission begins after a long delay
2. BNo emission occurs at any intensity
3. CEmission occurs with zero kinetic energy
4. DEmission begins once the intensity is high enough
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Answer: B · No emission occurs at any intensity
Each photon has energy hν, and one electron absorbs one photon. Below threshold, hν is less than the work function, so no electron can escape, however many photons arrive.
- A: Emission is instantaneous when it occurs, and here it does not occur at all.
- C: An electron cannot escape without energy at least equal to the work function.
- D: More photons do not help, as one photon gives its energy to one electron.
Remember this
Emission depends on frequency, not intensity; above threshold, intensity sets the number of electrons, frequency their energy.
Question 5 · Easy
Photons of energy 5.0 eV strike a metal of work function 2.0 eV. What stopping potential do the emitted electrons have?
1. A7.0 V
2. B5.0 V
3. C3.0 V
4. D2.5 V
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Answer: C · 3.0 V
Einstein's equation gives K_max = hν - φ = 5.0 - 2.0 = 3.0 eV. The stopping potential satisfies eV₀ = K_max, so V₀ = 3.0 V.
- A: The work function was added instead of subtracted.
- B: The work function was ignored.
- D: The photon energy was divided by the work function.
Remember this
eV₀ = hν - φ, so the stopping potential in volts equals the maximum kinetic energy in eV.
Question 6 · Medium
Assertion A: Graphs of stopping potential against frequency for two different metals are parallel straight lines. Reason R: The slope of each graph equals the work function of the metal divided by e. Choose the correct answer.
1. ABoth A and R are correct but R is not the correct explanation of A
2. BA is not correct but R is correct
3. CBoth A and R are correct and R is the correct explanation of A
4. DA is correct but R is not correct
Answer and explanation in LearnFree account
Question 7 · Medium
Stopping potential is plotted against frequency for two metals P and Q. If light of frequency 10 × 10¹⁴ Hz falls on each, what is the ratio of the maximum kinetic energy of photoelectrons from P to that from Q?
Figure: MedExams
1. A3 : 2
2. B1 : 1
3. C5 : 3
4. D2 : 3
Answer and explanation in LearnFree account
Question 8 · Easy
Maximum kinetic energy of photoelectrons from a metal is plotted against the frequency of the incident light. Using h = 6.6 × 10⁻³⁴ J s, what is the work function of the metal?
Figure: MedExams
1. A9.9 × 10⁻¹⁹ J
2. B5.0 × 10⁻¹⁹ J
3. C3.3 × 10⁻¹⁹ J
4. D6.6 × 10⁻¹⁹ J
Answer and explanation in LearnFree account
Question 9 · Medium
A photocell is lit by light of one frequency at two intensities, I₁ > I₂. Which graph correctly shows photocurrent i against anode potential V for both?
Figure: MedExams
1. AGraph A
2. BGraph C
3. CGraph D
4. DGraph B
Answer and explanation in LearnFree account
Question 10 · Hard
A proton (mass m, charge e), a deuteron (mass 2m, charge e) and an alpha particle (mass 4m, charge 2e) are each accelerated from rest through a potential difference V. Lines P, Q and R plot their de Broglie wavelengths against 1/√V. Identify the lines.
Figure: MedExams
1. AP: proton, Q: alpha particle, R: deuteron
2. BP: alpha particle, Q: deuteron, R: proton
3. CP: deuteron, Q: proton, R: alpha particle
4. DP: proton, Q: deuteron, R: alpha particle
Answer and explanation in LearnFree account

Dual Nature of Radiation & Matter

A deuteron and an α-particle, both initially at rest, are accelerated through the same potential difference. What is the ratio of the de Broglie wavelength of the deuteron to that of the α-particle?

A metal is lit in turn by light of wavelength 200 nm and of 400 nm. The stopping potential with 200 nm is three times that with 400 nm. Taking hc = 1240 eV nm, find the work function of the metal.

In a photoelectric experiment, the frequency of light is kept fixed above threshold while its intensity is doubled. What happens to the saturation current and the stopping potential?

Light below a metal's threshold frequency falls on it, and its intensity is increased many times. What happens to photoelectric emission?

Photons of energy 5.0 eV strike a metal of work function 2.0 eV. What stopping potential do the emitted electrons have?

Assertion A: Graphs of stopping potential against frequency for two different metals are parallel straight lines. Reason R: The slope of each graph equals the work function of the metal divided by e. Choose the correct answer.

Stopping potential is plotted against frequency for two metals P and Q. If light of frequency 10 × 10¹⁴ Hz falls on each, what is the ratio of the maximum kinetic energy of photoelectrons from P to that from Q?

Maximum kinetic energy of photoelectrons from a metal is plotted against the frequency of the incident light. Using h = 6.6 × 10⁻³⁴ J s, what is the work function of the metal?

A photocell is lit by light of one frequency at two intensities, I₁ > I₂. Which graph correctly shows photocurrent i against anode potential V for both?

A proton (mass m, charge e), a deuteron (mass 2m, charge e) and an alpha particle (mass 4m, charge 2e) are each accelerated from rest through a potential difference V. Lines P, Q and R plot their de Broglie wavelengths against 1/√V. Identify the lines.