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 $J s$ ) 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, $( KE )_{\max }=\frac{ hc }{\lambda}-\phi= eV _{0}$
$\frac{ hc }{\lambda}=\phi+ eV _{0}$
$\frac{ hc }{3 \times 10^{-7}}=\phi+ eV _{0}$
$\frac{ h \times 3 \times 10^{8}}{3 \times 10^{-7}}=\phi+1.6 \times 10^{-19} \times 2$ $h \times 10^{15}=\phi+3.2 \times 10^{-19} \ldots$(i)
$\frac{ h \times 3 \times 10^{ 8 }}{4 \times 10^{-7}}=\phi+1.6 \times 10^{-19} \times 1$ $0.75 h \times 10^{15}=\phi+1.6 \times 10^{-19} \ldots$(ii)
by equation (i) $-( ii )$
$0.25 h \times 10^{15}=1.6 \times 10^{-19}$
$h =\frac{1.6 \times 10^{-34}}{0.25}$
$h =6.4 \times 10^{-34}$