Question 6 of 10intermediate🔧 ApplyNumerical3 marks

The threshold frequency for a certain metal is 3.3×3.3 \times 10^{14} Hz. If light of frequency 8.2×8.2 \times 10^{14}$$ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Correct Answer

The cut-off voltage for the photoelectric emission is approximately 2.03 V.

Exercise: EXERCISES | Q: 11.6 | (Page 285)
For More Understanding

Explanation

This numerical problem applies Einstein's photoelectric equation. The context establishes that the maximum kinetic energy of photoelectrons equals the energy of incident photons minus the work function. The work function can be expressed in terms of threshold frequency. The stopping potential (cut-off voltage) is directly related to the maximum kinetic energy of the emitted photoelectrons.

Solution Steps

  1. Step 1: Identify given values — Threshold frequency ν₀ = 3.3×103.3 \times 10¹⁴ Hz, Incident frequency ν = 8.2×108.2 \times 10¹⁴ Hz, Planck's constant h = 6.63×106.63 \times 10⁻³⁴ J s, Charge of electron e = 1.6×101.6 \times 10⁻¹⁹ C.

  2. Step 2: Apply Einstein's photoelectric equation — The maximum kinetic energy is K_max = hν – hν₀ = h(ν – ν₀).

  3. Step 3: Relate kinetic energy to stopping potential — The stopping potential V₀ satisfies K_max = eV₀, therefore eV₀ = h(ν – ν₀).

  4. Step 4: Calculate the frequency difference — ν – ν₀ = (8.2 – 3.3) × 10¹⁴ = 4.9×104.9 \times 10¹⁴ Hz.

  5. Step 5: Compute the cut-off voltage — V₀ = h(ν – ν₀)/e = (6.63×106.63 \times 10⁻³⁴ × 4.9×104.9 \times 10¹⁴)/(1.6×101.6 \times 10⁻¹⁹) = 32.49×1032.49 \times 10⁻²⁰/1.6×101.6 \times 10⁻¹⁹ = 2.03 V (approximately).

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