Question

A stopping potential of $0.82\text{V}$ is required to stop the emission of photoelectrons from the surface of a metal by light of wavelength $4000\dot{\text{A}}$. For light of wavelength $3000\dot{\text{A}}$, the stopping potential is $1.85\text{V}$. If the value of Planck's constant is $6.6\times {10}^{-x}$. What is the value of $x$?

• A. $10$
• B. $17$
• C. $34$
• D. $20$

Answer 1

The correct option is C $34$

The formula for stopping potential is given by,

$e{V}_0=\frac{hc}{\lambda }-\varphi$

So for same $h$ and $\varphi$,

$\frac{hc}{{\lambda }_1}-e{V}_{01}=\frac{hc}{{\lambda }_2}-e{V}_{02}$

$\Rightarrow hc(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})=e(V_{01}-V_{02})$

Substituting the corresponding values,

$h\times 3\times {10}^8(\frac{1}{3000\times {10}^{-10}}-\frac{1}{4000\times {10}^{-10}})=1.6\times {10}^{-19}(1.85-0.82)$

$\Rightarrow h\approx 6.6\times {10}^{-34}\text{Js}$

Given, $h=6.6\times {10}^{-x}\text{Js}$

$\therefore x=34$

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