Question

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectric energies were measured by applying a stopping potential. The relevant data for the wavelength ($\lambda$) of incident light at the corresponding stopping potential ($V_0$) are given below:

$\lambda \left(\mu m\right)$	$V_0$ (Volt)
$0.3$	$2.0$
$0.4$	$1.0$
$0.5$	$0.4$

Given, $c=3\times {10}^{8}m{s}^{-1}$ and $e=1.6\times {10}^{-19}C$, Planck's constant (in units of $J-s$) from such an experiment is

• A. $6.0\times {10}^{-34}$
• B. $6.4\times {10}^{-34}$
• C. $6.6\times {10}^{-34}$
• D. $6.8\times {10}^{-34}$

Answer 1

The correct option is B $6.4\times {10}^{-34}$

Einstein's Photoelectric equation
$\frac{hc}{\lambda }-\varphi =e{V}_0$
From the given data;
for first wavelenght$\frac{hc}{0.3\times {10}^{-6}}-\varphi =2e$ .....(1)

for secound wavelenght$\frac{hc}{0.4\times {10}^{-6}}-\varphi =e$ .....(2)
Subtracting (1) to (2), we get,
$\Rightarrow hc(\frac{1}{0.3}-\frac{1}{0.4}){10}^6=e$
where, $c=3\times {10}^{8}m{s}^{-1}$ and $e=1.6\times {10}^{-19}C$
$\Rightarrow h\times (3\times {10}^8)(\frac{1}{0.3}-\frac{1}{0.4}){10}^6=1.6\times {10}^{-19}$
$h=6.4\times {10}^{-34}Js$