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Question:

The photoelectric threshold wavelength of silver is 3250 × 10⁻¹⁰ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10⁻¹⁰ m is :(Given h = 4.14 × 10⁻¹⁵ eVs and c = 3 × 10⁸ ms⁻¹)

≈6 × 10⁶ ms⁻¹

≈61 × 10³ ms⁻¹

≈0.3 × 10⁶ ms⁻¹

≈6 × 10⁵ ms⁻¹

Solution:

λ₀ = 3250 × 10⁻¹⁰ m = 3250 Å
λ = 2536 × 10⁻¹⁰ m = 2536 Å
1/2mv² = hc[(1/λ) - (1/λ₀)]
v = √(2hcm[(1/λ) - (1/λ₀)])
v = √(2 × 12400 × 1.6 × 10⁻¹⁹ × 9.1 × 10⁻³¹[(1/2536) - (1/3250)])
v = 0.6 × 10⁶ m/s = 6 × 10⁵ m/s
Hence, option (D) is correct answer.