Question

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

• A. $6\times {10}^{-5}V{s}^{-1}$
  
• B. $2\times {10}^{-15}V{s}^{-1}$
  
• C. $4\times {10}^{-55}V{s}^{-1}$
  
• D. $4\times {10}^{-15}V{s}^{-1}$
  

Answer 1

The correct option is D $4\times {10}^{-15}V{s}^{-1}$

In emission of electron, Potential Energy

$eV=\frac{hc}{\lambda }-W_0$

When light of wavelength $\lambda =0.6mm$and stopping potential $0.5V$
$0.5e=\frac{hc}{6\times {10}^{-7}}-W_0......\left(1\right)$

When light of wavelength $\lambda =0.4mm$and stopping potential $1.5V$
$1.5e=\frac{hc}{4\times {10}^{-7}}-W_0......\left(2\right)$
subtract equation (1) from (2) $e=\frac{hc}{{10}^{-7}}[\frac{1}{4}-\frac{1}{6}]$

$\Rightarrow \frac{h}{e}=\frac{12\times {10}^{-7}}{3\times {10}^8}=4\times {10}^{-15}V{s}^{-1}$