A photosensitve metallic surface has work fuction ` hv_0` If photons of ` 2hv_0` fall on the suface , the electrons come out with a maximum velocity of ` 4xx 10^6 m//s` . If photon energy is increases to ` 5hv_0` the maximum velocity of photoelectrons will be:

Options

A. ` 2xx 10^7 m//s`
B. `82xx 10^6 m//s`
C. ` 2xx 10^76m//s`
D. ` 8xx 10^5 m//s`

Answer

B

Explanation

` 2h-0 = hv_0 + K.E._1` ,
or `hv_0 = K.E. _1 = 1/2 mu_1^2`
Now `5 hv_0= hv_0 + K.E._2`
`K. E._2 = 4 hv_0 = 1/2 mu_1=2^2`
` :. U_2^2 /u_1^2 =4`
or ` u_2 = 2 u_1` .