The critical frequency for emitting photoellectrons from a metal surface is ` 5 xx 10^(14) sec ^(-1)`. What should be the frequency fo radiation to produce photoelectons having twice the kinetic energy of those produced by the radiation fo frequency ` 10^(15) sec ^(-1)` ?

Explanation

Given critical frequency `= 5 xx 10^(14) sec^(-1)`
When surface is exposed to frequency of ` 10^(15) sec^(-1)`
Energy fo photon ` = hv_0 + K.E.`
` h xx 10^(15) = h xx 5 xx 10^(14) + K.E`
` therefore K.E. = 10 xx h xx 10^(14)`
`= 5 xx h xx 10^(14)`
` = 5 xx h xx 10^(14)`
Now for (II) case . if K.E. becomes twice of this then New
` K.E, = 2 xx 5 xx h xx 10^(14)`
The energy of photon ` = hv_0 + K. E`
` hv= 5xx 10^(14) xx h + 2 xx 5 xx h xx 10^(14)`
` therefore v= 15 xx 10 ^(14) sec^(-1)` .