Question

Find the slope of the graph shown in the figure.


Where:
$V_s\to$ stopping potential $\left(\text{V}\right)$
$f\to$ frequency of the incident radiation $\left(\text{Hz}\right)$

• A. $he$
• B. $\frac{hc}{e}$
• C. $\frac{h}{e}$
• D. $h$

Answer 1

The correct option is C $\frac{h}{e}$

According to Einstein's photoelectric equation,

${\left(KE\right)}_{max}=E-{\varphi }_0$

If $V_s$ is the stopping potential in $\text{V}$, then

$e{V}_s=hf-h{f}_0$

$\Rightarrow V_s=\frac{h}{e}f-\frac{h{f}_0}{e}$

Comparing the above equation with the equation of straight line, $y=mx+c$,we get

$\text{Slope},m=\frac{h}{e}$

So, the slope of the given graph represents $h/e$.

Hence, option $\left(\text{B}\right)$ is correct.