If the frequency of light in a photoelectric experiment is doubled the stopping potential will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → becomes more than the double
According to Einstein's Photoelectric equation -
$K.E._{max}=hv-\phi$ ...(1)
and,
$K.E._{max}=eV_s$ [$V_s$ = Stopping potential] ...(2)
from (1) and (2) we get,
$V_0=\frac{hv-\phi}{e}$
if the frequency of the light is doubled,
$V_0'=\frac{h(2v)-\phi}{e}=2hv-\phi$
$V_0'=2V_0+\frac{\phi}{e}$
Since, $\frac{\phi}{e}$ is constant for a given material, the stopping potential becomes slightly more than double.