Question

Light of wavelength $0.6\mu m$ 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 $0.4\mu m$ from a sodium lamp, the stopping potential is $1.5V$, with this data, the value of $h/e$ is

• A. $4\times 10^{-19}Vs$
• B. $0.25\times 10^{-15}Vs$
• C. $4\times 10^{-15}Vs$
• D. $4\times 10^{-8}Vs$

Answer 1

Answer: (C)

According to photoelectric equation
$eV=\frac{hc}{\lambda }-W_0$
$0.5e=\frac{hc}{6\times 10^{-7}}-W_0$
$\Rightarrow 0.5=\frac{h}{e}\times \frac{c}{6\times 10^{-7}}-\frac{W_0}{e}$ ...(i)
Similarly,
$1.5=\frac{h}{e}\times \frac{c}{4\times 10^{-7}}-\frac{W_0}{e}$ ...(ii)
From Eqs. (i) and (ii),
$1=\frac{h}{e}\times \frac{c}{10^{-7}}\left(\frac{1}{4}-\frac{1}{6}\right)$
$\Rightarrow \frac{h}{e}=\frac{12\times 10^{-7}}{3\times 10^8}$
$=4\times 10^{-15}$