Maximum kinetic energy of photoelectrons from a metal
surface is `K_0` when wavelngth of incident light is `lambda`. If wavelength is decreased
to `lambda|2`, the maximum kinetic energy of photoelectrons becomes
(a) `=2K_0` (b)`gt2K_0` (c ) `lt2K_0`

Answer

B

Explanation

Using the equation,
`K_(max) = E - W` ,we have
`K_0 =(hc)/ (lambda) -W
with wavelength `(lambda)/(2)`, suppose the maximum kinetic energy is `K'_0`, then
`K'_0 =(hc)/(lambda//2) -W`
= 2(hc)/(lambda) - W`
but `(hc)/lambda) - w is `K_0`. Therefore, `K_0 = 2k_0+W`
or `K'_0 gt 2K_0` `:.` The correct option is (b).