Question

Given figure shows few data points in a photo-electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Planck’s constant $h=6.62\times {10}^{-34}J.s$)

• A. $2.10eV$
• B. $2.27eV$
• C. $2.59eV$
• D. $1.93eV$

Answer 1

The correct option is B

$2.27eV$

Step 1. Given data

Planck constant, $h=6.62\times {10}^{-34}J.s$

From diagram we have threshold frequency, $f_0=5.5\times {10}^{14}Hz$

Step 2. Finding threshold energy

The point $A$ in graph is point represents that the voltage $V$ with negative stopping potential.

The point $B$ represents the point when no voltage is applied and the frequency corresponds to threshold frequency $f=5.5\times {10}^{14}Hz$.

The point $C$ represents the stopping potential of $V$

The minimum amount of energy required to remove an electron from the metal is called the work function which is given by $\phi =h{f}_0$

$$\begin{gathered} \Rightarrow \phi =6.62\times {10}^{-34}\times 5.5\times {10}^{14} \\ \Rightarrow \phi =36.41\times {10}^{-20}J \\ \Rightarrow \phi =\frac{36.41\times {10}^{-20}}{1.6\times {10}^{-19}}eV \\ \Rightarrow \phi =2.27eV \end{gathered}$$

Hence, option B is correct