Question

When the incident frequency is $f_0,K$ is the $(KE{)}_{max}$ of the electrons emitted and $\varphi$ is the work function of the surface. If incident frequency is doubled, new $(KE{)}_{max}$ will be

• A. 2 $K$
• B. 2 $K-\varphi$
• C. 2 $K+\varphi$
• D. 2 $K-2\varphi$

Answer 1

The correct option is C 2 $K+\varphi$

The incident frequency is $f_0,K$ is the $(KE{)}_{max}$ of the electrons emitted and $\varphi$ is work function of the surface.
Then,
$K=hf-\varphi$
Now, the incident frequency is doubled so the new $(KE{)}_{max}$ will be
$K^{\prime }=h\left(2f\right)-\varphi$
or new $(KE{)}_{max}=2hf-\varphi =hf+K=hf-\varphi +\varphi +K=2K+\varphi$

So, the answer is option (C).