Radiations of frequencies `8xx10^(15)` Hz and `5xx10^(15)` Hz are incident one after the another on the same photosensitive surface. If is found that the kinetic energies of the photo electrons emitted from the surface are in the ratio of 2:1 . What is the threshold frequency for the surface ?

Options

A. `1.5 xx 10^(15)Hz`
B. `2 xx 10^(15)Hz`
C. `2.5 xx 10^(15)Hz`
D. `3 xx 10^(15)`Hz

Answer

B

Explanation

`h upsilon = h upsilon_0 + K. E(K)`
`therefore h xx 8 xx 10^(15) = hv_0 +2 K`
`h xx 5 xx 10^15 = h upsilon_0 + K`
`therefore 2(h xx 5 xx 10^(15))=2h upsilon +2 K`
Subtracting (1) from (2), we get
`h xx 10^(15)[10-8]=h upsilon_0`
`therefore upsilon= 2 xx 10^(15) Hz`