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 $0.5\, V$ with light of wavelength $0.4\, \mu m$ from a sodium lamp, the stopping potential is $1.5\, V$, 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
$e V=\frac{h c}{\lambda}-W_{0}$
$0.5\, e=\frac{h c}{6 \times 10^{-7}}-W_{0}$
$\Rightarrow 0.5=\frac{h}{e} \times \frac{c}{6 \times 10^{-7}}-\frac{W_{0}}{e}$ ...(i)
Similarly,
$1.5=\frac{h}{e} \times \frac{c}{4 \times 10^{-7}}-\frac{W_{0}}{e}$ ...(ii)
From Eqs. (i) and (ii),
$1=\frac{h}{e} \times \frac{c}{10^{-7}}\left(\frac{1}{4}-\frac{1}{6}\right)$
$\Rightarrow \frac{h}{e}=\frac{12 \times 10^{-7}}{3 \times 10^{8}}$
$=4 \times 10^{-15}$