Question

The stopping potential $V_0$ $\left(\text{in volts}\right)$ as a function of frequency $\left(\nu \right)$ for a sodium emitter, is shown in the figure. The work function of sodium, from the data given in the plot will be : $(\text{Given :}\text{Planck's constant}h=6.63\times {10}^{-34}\text{Js}\text{charge of an electron}|e|=1.6\times {10}^{-19}\text{C})$

• A. $1.82\text{eV}$
• B. $1.95\text{eV}$
• C. $1.66\text{eV}$
• D. $2.12\text{eV}$

Answer 1

The correct option is C $1.66\text{eV}$

The stopping potential will be zero when the incident frequency is equal to the threshold frequency.

Therefore, from the graph,

$\text{Threshold frequency},{\nu }_0=4\times {10}^{14}\text{Hz}$

$\Rightarrow \text{Work function}\varphi =h{\nu }_0$

$\Rightarrow \varphi =6.63\times {10}^{-34}\times (4\times {10}^{14})\text{J}$

$\Rightarrow \varphi =\frac{(6.63\times {10}^{-34})\times (4\times {10}^{14})}{1.6\times {10}^{-19}}\text{eV}$

$\Rightarrow \varphi =1.66\text{eV}$

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