Question 5 of 10beginner🔧 ApplyNumerical2 marks

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Correct Answer

Given: Cut-off voltage, V₀ = 1.5 V

Formula: The maximum kinetic energy of photoelectrons is related to the stopping potential by: K_max = eV₀

Calculation: K_max = e × V₀ K_max = e × 1.5 V K_max = 1.5 eV

Or in joules: K_max = 1.6×101.6 \times 10⁻¹⁹ C × 1.5 V K_max = 2.4×102.4 \times 10⁻¹⁹ J

Answer: The maximum kinetic energy of photoelectrons emitted is 1.5 eV (or 2.4×102.4 \times 10⁻¹⁹ J).

Exercise: EXERCISES | Q: 11.3 | (Chapter: 285)
For More Understanding

Explanation

This question tests the relationship between stopping potential and maximum kinetic energy of photoelectrons. The textbook context clearly states in Equation 11.1 that K_max = eV₀, where V₀ is the stopping potential (also called cut-off voltage). When the stopping potential is applied, it repels even the most energetic photoelectrons, bringing the photocurrent to zero. The kinetic energy in electron-volts numerically equals the stopping potential in volts, making this a straightforward calculation.

Solution Steps

  1. Step 1: Identify the given value - Cut-off voltage V₀ = 1.5 V

  2. Step 2: Apply the formula K_max = eV₀ from Equation 11.1

  3. Step 3: Calculate K_max = e × 1.5 V = 1.5 eV (or 2.4×102.4 \times 10⁻¹⁹ J)

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