When two monochromatic light of frequency and /2 are incident on a photoelectric metal, their stopping potential becomes V_s/2 and V_s respectively. T - Physics MCQ NEET 2022 2026 | ExamTest.live

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When two monochromatic light of frequency ν and ν /2 are incident on a photoelectric metal, their stopping potential becomes V_{s} /2 and V_{s} respectively. The threshold frequency for this metal is:

Solution & Step-by-step Explanation

Einstein's photoelectric equation: e V_{0} = h ν - h ν_{0} .Case 1: e (V_{s} /2) = h ν - h ν_{0} \dots (1) Case 2: e V_{s} = h (ν /2) - h ν_{0} \dots (2) Dividing (2) by (1): 2 = \frac{h ν /2 - h ν _{0}}{h ν - h ν _{0}} 2 h ν - 2 h ν_{0} = \frac{h ν}{2} - h ν_{0} 2 h ν - \frac{h ν}{2} = 2 h ν_{0} - h ν_{0} \frac{3 h ν}{2} = h ν_{0} ⟹ ν_{0} = \frac{3}{2} ν (Note: Using the specific values provided in the prompt text, though usually stopping potential increases with frequency).

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When two monochromatic light of frequency ν and ν /2 are incident on a photoelectric metal, their stopping potential becomes V_{s} /2 and V_{s} respectively. The threshold frequency for this metal is:

2 ν

3 ν

2/3 ν

3/2 ν

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