Question

The frequency of incident light falling on a photosensitive metal plate is doubled, the K.E of the emitted photo-electrons is
A. Double the earlier value
B. Unchanged
C. More than doubled
D. Less than doubled

Answer 1

Correct Answer - C
`E = W_(0) + K_(max) rArr K_(max) = E - W_(0) = hv - W_(0)`
`rArr K_(1) = hv - W_(0) ` and `K_(1) = 2 hv - W_(0) rArr K_(1) gt 2K_(1)`