Question

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

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

Given that $c=3\times 10^{8}m{s}^{-1}$ and $e=1.6\times 10^{-19}C$, Planck's constant (in units of $Js$ ) found 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

Answer: (B)

We have, $(KE{)}_{\max }=\frac{hc}{\lambda }-\varphi =e{V}_0$
$\frac{hc}{\lambda }=\varphi +e{V}_0$
$\frac{hc}{3\times 10^{-7}}=\varphi +e{V}_0$
$\frac{h\times 3\times 10^8}{3\times 10^{-7}}=\varphi +1.6\times 10^{-19}\times 2$ $h\times 10^{15}=\varphi +3.2\times 10^{-19}\dots$(i)
$\frac{h\times 3\times 10^8}{4\times 10^{-7}}=\varphi +1.6\times 10^{-19}\times 1$ $0.75h\times 10^{15}=\varphi +1.6\times 10^{-19}\dots$(ii)
by equation (i) $-(ii)$
$0.25h\times 10^{15}=1.6\times 10^{-19}$
$h=\frac{1.6\times 10^{-34}}{0.25}$
$h=6.4\times 10^{-34}$