Question

Light of wavelength $0.6mm$ from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is $0.5V$. With light of wavelength $0.4mm$ from a sodium lamp, the stopping potential is $1.5V$. With this data, the value of h/e is

• A. $4\times 10^{-59}Vs$
• B. $0.25\times 10^{15}Vs$
• C. $4\times 10^{-15}Vs$
• D. $4\times 10^{-8}Vs$

Answer 1

Answer: (C)

Here, $eV=\frac{hc}{\lambda }-W_0$
$0.5e=\frac{hc}{6\times 10^{-7}}-W_0$
$\Rightarrow 0.5=\frac{h}{e}\left(\frac{c}{6\times 10^{-7}}\right)-\frac{W_0}{e}\dots \left(i\right)$
Similarly, $1.5=\frac{h}{e}\left(\frac{c}{4\times 10^{-7}}\right)-\frac{W_0}{e}\dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$,
$1=\frac{h}{e}\frac{c}{10^{-7}}\left[\frac{1}{4}-\frac{1}{6}\right]$
$\Rightarrow \frac{h}{e}=\frac{12\times 10^{-7}}{3\times 10^8}$
$=4\times 10^{-15}Vs$