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Hertz/Lenard observations and Einstein photoelectric equation: NEET MCQs with answers

90 questions on this topic of Dual Nature of Matter and Radiation (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Light whose photons carry 7 eV is incident on a metal with work function 2.5 eV. The maximum kinetic energy of the photoelectrons is:

  1. 4.5 eV
  2. 9.5 eV
  3. 7 eV
  4. 2.5 eV
Show answer and explanation

Answer: (A) 4.5 eV

Einstein’s photoelectric equation: K_max = hν − φ₀ = 7 − 2.5 = 4.5 eV.

Quick tip: K_max = hν − φ₀. The work function is subtracted, never added.

NCERT Class 12, Chapter 11

medium

2. Light of wavelength 300 nm falls on a metal with work function 3 eV. Taking hc = 1240 eV nm, the maximum kinetic energy of the photoelectrons is:

  1. 4.13 eV
  2. 7.13 eV
  3. 3 eV
  4. 1.13 eV
Show answer and explanation

Answer: (D) 1.13 eV

Einstein's equation: K_max = hc/λ − φ = 1240/300 − 3 = 4.13 − 3 = 1.13 eV.

hard

3. When light of frequency 3ν₀ is incident on a photosensitive metal plate of threshold frequency ν₀, the emitted electrons enter a region of uniform magnetic field B perpendicular to their velocity. If the maximum radius of the circular path traced by the electrons is R, what will be the maximum radius of their trajectory when the incident light frequency is increased to 9ν₀?

  1. 2 R
  2. 3 R
  3. √3 R
  4. 4 R
Show answer and explanation

Answer: (A) 2 R

The radius of an electron's circular path in a transverse magnetic field is R = mv / (qB) = √(2m K_max) / (eB). In the first case, K₁ = h(3ν₀ - ν₀) = 2hν₀, so R₁ = √(2m · 2hν₀) / (eB) = R. In the second case, K₂ = h(9ν₀ - ν₀) = 8hν₀. Since K₂ = 4K₁, the new radius is R₂ = √(2m · 8hν₀) / (eB) = √4 × R₁ = 2R.

medium

4. Consider the following two statements regarding Lenard and Hallwachs' observations: Statement I: The saturation photocurrent is directly proportional to the intensity of incident radiation of a given frequency. Statement II: The stopping potential depends linearly on the intensity of the incident radiation. Choose the correct option:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect.

Statement I is correct because the number of photoelectrons emitted per second is directly proportional to the incident light intensity. Statement II is incorrect because stopping potential depends only on the frequency of the incident radiation and the work function of the metal, and is independent of intensity.

easy

5. Photons of energy 3.5 eV fall on a metal surface of work function 2 eV. The maximum kinetic energy of the emitted photoelectrons is:

  1. 3.5 eV
  2. 1.5 eV
  3. 2 eV
  4. 5.5 eV
Show answer and explanation

Answer: (B) 1.5 eV

Einstein’s photoelectric equation: K_max = hν − φ₀ = 3.5 − 2 = 1.5 eV.

Quick tip: K_max = hν − φ₀. The work function is subtracted, never added.

NCERT Class 12, Chapter 11

hard

6. Two different photosensitive metals X and Y have work functions Φ_X = 2.0 eV and Φ_Y = 4.5 eV respectively. When both are irradiated with light of frequency ν, the maximum speed of photoelectrons from metal X is twice the maximum speed of photoelectrons from metal Y. The energy of the incident photon (hν) is:

  1. 6.50 eV
  2. 5.33 eV
  3. 4.80 eV
  4. 7.25 eV
Show answer and explanation

Answer: (B) 5.33 eV

Let E = hν. The maximum kinetic energy is K = (1/2)m v_max². Since (v_max)_X = 2 (v_max)_Y, we have K_X = 4 K_Y. Using Einstein's equation: E - 2.0 = 4(E - 4.5). This gives E - 2.0 = 4E - 18.0, so 3E = 16.0 eV, which yields E = 16/3 ≈ 5.33 eV.

easy

7. A metallic surface is irradiated with light of frequency ν = 8.0 × 10¹⁴ Hz. If the threshold frequency of the metal is ν₀ = 5.0 × 10¹⁴ Hz, what is the maximum kinetic energy of the emitted photoelectrons? (Take Planck's constant h = 6.63 × 10⁻³⁴ J s and 1 eV = 1.6 × 10⁻¹⁹ J)

  1. 0.62 eV
  2. 2.48 eV
  3. 1.24 eV
  4. 1.86 eV
Show answer and explanation

Answer: (C) 1.24 eV

Using Einstein's photoelectric equation: K_max = h(ν - ν₀) = (6.63 × 10⁻³⁴ J s)(8.0 × 10¹⁴ - 5.0 × 10¹⁴ s⁻¹) = 6.63 × 10⁻³⁴ × 3.0 × 10¹⁴ J = 1.989 × 10⁻¹⁹ J. Converting to eV: (1.989 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) ≈ 1.24 eV.

easy

8. A point source of monochromatic light is placed at a distance of 0.5 m from a photosensitive surface. The stopping potential is found to be V₀ and the saturation current is I₀. If the distance of the light source is doubled to 1.0 m, the new stopping potential and saturation current will be respectively:

  1. V₀/2 and I₀/2
  2. V₀ and I₀/4
  3. V₀/4 and I₀/4
  4. V₀ and I₀/2
Show answer and explanation

Answer: (B) V₀ and I₀/4

Intensity from a point source varies inversely as the square of the distance (I ∝ 1/r²), so doubling the distance reduces intensity to one-fourth, making the saturation current I₀/4. However, the frequency of light remains unchanged, so the stopping potential V₀ remains unchanged.

easy

9. In a photoelectric experiment, a graph of stopping potential (V₀) versus frequency (ν) of incident radiation is plotted. The ratio of the Planck constant to the elementary charge (h/e) is given by the:

  1. Area under the graph
  2. Intercept on the frequency axis
  3. Intercept on the potential axis
  4. Slope of the graph
Show answer and explanation

Answer: (D) Slope of the graph

From the equation eV₀ = hν - Φ, we have V₀ = (h/e)ν - (Φ/e). This is of the form y = mx + c, where the slope m = h/e.

easy

10. Photons of energy 6.5 eV fall on a metal surface of work function 2.5 eV. The maximum kinetic energy of the emitted photoelectrons is:

  1. 6.5 eV
  2. 2.5 eV
  3. 4 eV
  4. 9 eV
Show answer and explanation

Answer: (C) 4 eV

Einstein’s photoelectric equation: K_max = hν − φ₀ = 6.5 − 2.5 = 4 eV.

Quick tip: K_max = hν − φ₀. The work function is subtracted, never added.

NCERT Class 12, Chapter 11

easy

11. Light whose photons carry 6.5 eV is incident on a metal with work function 2 eV. The maximum kinetic energy of the photoelectrons is:

  1. 8.5 eV
  2. 6.5 eV
  3. 4.5 eV
  4. 2 eV
Show answer and explanation

Answer: (C) 4.5 eV

Einstein’s photoelectric equation: K_max = hν − φ₀ = 6.5 − 2 = 4.5 eV.

Quick tip: K_max = hν − φ₀. The work function is subtracted, never added.

NCERT Class 12, Chapter 11

easy

12. Philipp Lenard observed that when the accelerating potential applied to the collector plate in a photoelectric tube is increased beyond a certain positive value, the photocurrent:

  1. Keeps increasing linearly without bound
  2. Increases exponentially with collector potential
  3. Suddenly drops to zero due to electron-electron collisions
  4. Becomes constant and reaches a saturation value
Show answer and explanation

Answer: (D) Becomes constant and reaches a saturation value

Lenard observed that as the accelerating potential of the collector plate is increased, more emitted electrons are collected until all electrons emitted per second reach the plate. Beyond this point, the photocurrent attains a maximum constant value known as the saturation current.

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