Question

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

• A. $1.95eV$
• B. $1.82eV$
• C. $1.66eV$
• D. $2.12eV$

Answer 1

The correct option is C $1.66eV$

$hv=\varphi +e{v}_0$
$v_0=\frac{hv}{e}-\frac{\varphi }{e}$
$v_0$ is zero fro $v=4\times {10}^{14}Hz$
$0=\frac{hv}{e}-\frac{\varphi }{e}\Rightarrow \varphi =hv$
$=\frac{6.63\times {10}^{-34}\times 4\times {10}^{14}}{1.6\times {10}^{-19}}=1.66eV$.