Question

The threshold frequency of the metal of the cathode in a photoelectric cell is `1xx10^(15)Hz`. When a certain beamm of light is incident on the cathode, it is found that a stopping potential 4.144 V is required to reduce the current to zero. The frequency of the incident radiation is
A. `2.5xx10^(15)Hz`
B. `2xx10^(15)Hz`
C. `4.144xx10^(15)Hz`
D. `3xx10^(16)Hz`

Answer 1

Correct Answer - B
`V_(0)=(h)/(e)(v-v_(0))`
or `(v-v_(0))=(V_(0)xxe)/(h)=(4.144xx1.6xx10^(-19))/(6.63xx10^(-34))`
`v-1xx10^(15)=10^(15)`
frequency `=10^(15)+1xx10^(15)=2xx10^(15)Hz`