Question

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

• A. doubled
• B. halved
• C. increased but more than doubled of the previous KE
• D. remains unchanged
  

Answer 1

Answer: (C)

increased but more than doubled of the previous KE

$h{v}_1=h{v}_0+K{E}_1$ ---(1)

$2h{v}_1=h{v}_0+K{E}_2$ -----(2)

By dividing equations (1) and (2),
we get,

$K{E}_2=2K{E}_1+h{v}_0$

The value of kinetic energy will increase more than double of the previous $K.E.$

Hence, the correct option is C