When the frequency of light incident an a metallic plate is doubled , the KE of the emitted photoelectrons will be :

Options

A. doubled
B. halved
C. more than double
D. increases but less than double.

Answer

C

Explanation

Kinetic energy (K.E.) of a photoelectron `=h(v-v_(0))`
v=frequency of incident light
`"K.E."_(1)=h(v-v_(0))`
`"K.E."_(2)=h(v_(2)-v_(0))" "becausev_(2)=2v`
`"K.E."_(2)=h(2v-v_(0))=2h(v-(v_(0))/(2))`
`("K.E."_(2))/("K.E."_(1))=2((v-(v_(0)//2))/(v-v_(0)))`
As `v-(v_(0)//2)gtv-v_(0)`, thus new kinetic energy will be more than double.