The threshold frequency for certain metal is v0. When light of frequency 2v0 is incident on it, the maximum velocity of photoelectrons is 4 x 106 ms-1. If the frequency ofincident radiation is increased to 5v0, then the maximum velocity of photoelectrons will be
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a
4/5×106ms−1
b
2×106ms−1
c
8×106ms−1
d
2×107ms−1
answer is C.
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Detailed Solution
In the first case,12mv1max2=2hv0−hv0=hv0In the second case12mv2max2=5hv0−hv0=4hv0 Clearly, v2max is double of v1max ⇒v2max=2v1max=8×106ms−1