A light of frequency `1.6xx10^(16)` Hz when falls on a metal plate emits electrons that have double the kinetic energy compared to the kinetic energy of emitted electrons when frequency of `1.0xx10^(16)` Hz falls on the same plate. The threshold frequency `(lamda_(0))` of the metal in Hz is

Options

A. `1xx10^(15)`
B. `4xx10^(15)`
C. `3xx10^(15)`
D. `4xx10^(13)`

Answer

B

Explanation

When `1.6 xx 10^(16)` Hz frequency falls on metal plate than double kinetic energy obtained `h(1 cdot 6xx10^(16) -v_(0))=2 K. E rarr Eq. (i)`
When `1.0xx10^(16)` Hz frequency falls on same plate than K.E.
`h(1.0xx10^(16)-v_(0))=K.E rarr (Eq. (ii))`
From Eqs. (i) and (ii)
`(h(1.6 xx 10^(16)-v_(0)))/(h(1.0xx10^(16)-v_(0)))=(2 K.E)/(K.E)`
`1.6xx10^(16)-v_(0) =2xx1.0xx10^(16)-v_(0)`
`1.6xx10^(16) -2(1.0xx10^(16))=4xx10^(15)Hz`
`v_(0) =4xx10^(15) Hz`