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If the frequency of incident light is doubled while the intensity is halved, the number of photoelectrons emitted per second
A{'text': 'Doubles', 'label': 'A'}
B{'text': 'Halves', 'label': 'B'}
C{'text': 'Remains the same', 'label': 'C'}
D{'text': 'Depends on the metal only', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': 'Halves', 'label': 'B'}
1. Number of photons per second in the beam = Intensity / (h × frequency), approximately.
2. Doubling ν halves the energy each photon carries, so at the same power more photons per second.
3. But intensity is also halved, cancelling the frequency doubling effect: new N ~ (I/2) / (2ν) = (1/4) × original.
4. Actually simplifying: photon number scales like I/ν; halving I and doubling ν gives 1/4 as many. Closest to "halves" from listed options — with the caveat that above threshold N = number of photoelectrons per second is limited by intensity chiefly; N drops.
_Source: NCERT Class 12 Physics, Ch 11 "Dual Nature of Radiation and Matter", §11.4.1_
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