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. $4\times 10^{-59}\,V\,s$
• B. $0.25 \times 10^{15}\,V\,s$
• C. $4 \times 10^{-15}\,V\,s$
• D. $4 \times 10^{-8}\,V\,s$

Answer 1

Answer: (C)

Here, $eV =\frac{hc}{\lambda}-W_{0}$
$0.5e=\frac{hc}{6\times10^{-7}}-W_{0}$
$\Rightarrow 0.5=\frac{h}{e}\left(\frac{c}{6\times10^{-7}}\right)-\frac{W_{0}}{e} \ldots\left(i\right)$
Similarly, $1.5=\frac{h}{e} \left(\frac{c}{4\times10^{-7}}\right)-\frac{W_{0}}{e} \ldots\left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$,
$1=\frac{h}{e} \frac{c}{10^{-7}} \left[\frac{1}{4}-\frac{1}{6}\right]$
$\Rightarrow \frac{h}{e}=\frac{12\times10^{-7}}{3\times10^{8}}$
$=4\times10^{-15}\,V\,s$