Question

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $\left(v_1>v_2\right)$. If the maximum values of kinetic energy of the photo-electrons emitted in the two cases are in ratio of $1:k$, then the threshold frequency of the metallic surface is:

• A. $\frac{v_{1}k-v_2}{k-1}$
• B. $\frac{v_{2}k-v_1}{k-1}$
• C. $\frac{v_{1}k-v_2}{1-k}$
• D. $\frac{v_1-v_{2}k}{k-1}$

Answer 1

The correct option is A

$\frac{v_{1}k-v_2}{k-1}$

Step 1: Given

Frequencies are $v_1$ and $v_2$, such that, $v_1>v_2$

Ratio of kinetic energies: $\frac{K{E}_1}{K{E}_2}=\frac{1}{k}$, where $K{E}_1$ is the kinetic energy for frequency $v_1$ and $K{E}_2$ is the kinetic energy for frequency $v_2$

Step 2: Formula Used

1. The photoelectric equation is $hv=KE+\varphi$, where $h$ is the Planck's constant, $v$ is the frequency of incident light, $KE$ is the maximum kinetic energy of the electrons emitted and $\varphi$ is the work function of the metal.
2. And the formula of work function of a metal is given as $\varphi =h{\nu }_0$, where $v_0$ is the threshold frequency of the metal.
3. From $hv=KE+\varphi$ and $\varphi =h{\nu }_0$ it can be deduced that $KE=hv-h{v}_0\dots \left(1\right)$.

Step 3: Solution

Using equation $\left(1\right)$ the kinetic energies for frequencies $v_1$ and $v_2$ are $K{E}_1=h{v}_1-h{v}_0$ and $K{E}_2=h{v}_2-h{v}_0$.

Substitute the values in the equation $\frac{K{E}_1}{K{E}_2}=\frac{1}{k}$ and solve for threshold frequency.

$$\begin{gathered} \frac{h{v}_1-h{v}_0}{h{v}_2-h{v}_0}=\frac{1}{k} \\ \Rightarrow \frac{h\left(v_1-v_0\right)}{h\left(v_2-v_0\right)}=\frac{1}{k} \\ \Rightarrow k{v}_1-k{v}_0=v_2-v_0 \\ \Rightarrow k{v}_0-v_0=k{v}_1-v_2 \\ \Rightarrow v_0\left(k-1\right)=k{v}_1-v_2 \\ \Rightarrow v_0=\frac{k{v}_1-v_2}{k-1} \end{gathered}$$

Hence, option A is correct.