Question

Light of frequency $f$ is incident on a photosensitive surface of threshold frequency $\frac{f}{2}$. The energy of the fastest photoelectron is found to be equal to $K$. If the frequency of the incident radiation is doubled, then the energy of the fastest electron will be

• A. $3K$
• B. $2K$
• C. $K$
• D. $\frac{K}{2}$

Answer 1

The correct option is A $3K$

As per Einstein's photoelectric equation,

In case $1$ : $hf=\frac{hf}{2}+K$

Thus, $K=\frac{hf}{2}$

In case $2$ : $h\left(2f\right)=\frac{hf}{2}+K^{{}^{\prime }}$

Thus $K^{{}^{\prime }}=\frac{3hf}{2}=3K$