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PHY 104 PHY 104 Past Questions Objective Question
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Question

If a photo-emissive surface has a threshold wavelength of 0.65 µm, calculate the maximum speed of the electrons emitted by violet light of wavelength 0.40 µm.
Options
A 4.1 × 10⁵ m/s
Correct Answer
B 6.5 × 10⁵ m/s
C 10.4 × 10⁵ m/s
D 3.5 × 10⁵ m/s
Correct Answer

Option A is the correct answer.

Detailed Explanation

K_max = hc(1/λ - 1/λ₀) = 1240 eV·nm (1/400 - 1/650) = 1240(0.0025 - 0.001538) = 1240 × 0.000962 = 1.193 eV. K = ½mv² => v = √(2K/m) = √(2 × 1.193 × 1.602e-19 / 9.11e-31) = √(3.82e-19/9.11e-31) = √(4.194e11) = 6.477e5 m/s. None of the options match. If K = 1.193 eV, v = 6.48e5 m/s. The closest is 6.5 × 10⁵ m/s.

Hint

Use K_max = hc(1/λ - 1/λ₀).

Question Info

Course: PHY 104
Topic: PHY 104 Past Questions
Type: Objective
Difficulty: Not specified
Question 915 of 947
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