Question

In a photo emissive cell with exiting wavelength $\lambda$, the fastest electron has a speed v. If the exciting wavelength be changed to $\dfrac{3\lambda}{4}$ , then the speed of the fastest electron will be:
A. $v \left(\dfrac{3}{4}\right)^{\dfrac{1}{2}}$
B. $v \left(\dfrac{4}{3}\right)^{\dfrac{1}{2}}$
C. Less than $v \left(\dfrac{4}{3}\right)^{\dfrac{1}{2}}$
D. Greater than $v \left(\dfrac{4}{3}\right)^{\dfrac{1}{2}}$

Answer 1

Hint: We will use Einstein’s famous equation of photoelectric effect. For the two given cases, two different equations will be obtained. After solving them, we will try to find the relationship between the former maximum velocity and the later maximum velocity of the electron. Hence, the solution will be found.

Formula used: $\dfrac{1}{2}mv^2=\dfrac{hc}{\lambda}-W_0$

Complete step-by-step solution:
In the first case, the maximum velocity of the electron is v and the wavelength used is $\lambda$ . So, the Einstein’s equation of photoelectric effect is given by,
$\dfrac{1}{2}m{{v}^{2}}=\dfrac{hc}{\lambda }-{{W}_{0}}............(i)$
Here, h= Planck’s constant, c= velocity of light, $W_0$ is called work function and m= mass of an electron.
In the second case, let the velocity be $v_2$ . The wavelength is given to be $\dfrac{3\lambda}{4}$.
So, the second equation is given by,
$\dfrac{1}{2}mv_{2}^{2}=\dfrac{4hc}{3\lambda }-{{W}_{0}}.........(ii)$
Now, subtract the second equation from the first one. What we obtain is …
$\dfrac{1}{2}m\left(v_2^2-v^2\right)=hc\dfrac{4}{3\lambda}-\dfrac{1}{\lambda}=\dfrac{hc}{3\lambda}$
$\Rightarrow v_{2}^{2}-{{v}^{2}}=\dfrac{2hc}{3m\lambda }=\dfrac{2}{3m}.\dfrac{hc}{\lambda }$
Here, we will use the value of $\dfrac{hc}{\lambda}$ from the first equation. Then we get,
$\begin{aligned} & v_{2}^{2}-{{v}^{2}}=\dfrac{2}{3m}.\left( \dfrac{1}{2}m{{v}^{2}}+{{W}_{0}} \right) \\\ & v_{2}^{2}-{{v}^{2}}=\dfrac{1}{3}.\left( {{v}^{2}}+\dfrac{2{{W}_{0}}}{m} \right) \\\ & \Rightarrow v_{2}^{2}={{v}^{2}}+\dfrac{1}{3}{{v}^{2}}+\dfrac{2{{W}_{0}}}{3m} \\\ \end{aligned}$
Later we obtain,
$v_{2}^{2}=\dfrac{4{{v}^{2}}}{3}+\dfrac{2{{W}_{0}}}{3m}$
Hence, clearly $v_2^2$ is greater than $\dfrac{4v^2}{3}$. So, v is greater than $v \left(\dfrac{4}{3}\right)^{\dfrac{1}{2}}$.
Hence option D is the correct answer.
Additional information: Work function is defined as the minimum energy required by an external agent to emit an electron from the metal surface. If the incident light has less energy than the work function, photoemission won’t occur. The minimum frequency of incident light that can cause photoelectric effect is called the Threshold frequency. It is given by,
$\nu_0=\dfrac{W_0}{h}$
The corresponding wavelength to the threshold frequency is called the threshold wavelength.

Note: The photoelectric equation has many forms but one should use what is necessary to solve a particular problem. If the question says about threshold voltage, $V_s$, use the form,
$eV_s=\dfrac{hc}{\lambda}-W_0$
Here, e is the electronic charge.
All the terms like $eV_s$ or $\dfrac{1}{2}mv^2$ refers to the maximum kinetic energy of an electron.