Question

The work function of the nickel is $5\, eV$. When a light of wavelength $2000\,\mathring{A}$ falls on it, it emits photoelectrons in the circuit. Then the potential difference necessary to stop the fastest electrons emitted is (given $\left. h=6.67 \times 10^{-34} J - s \right)$

• A. 1.0 V
• B. 1.75 V
• C. 1.25 V
• D. 0.75 V

Answer 1

Answer: (C)

Stopping potential is given by
$e V_{0}=\frac{h c}{\lambda}-\phi$
$\therefore e V_{0}=\frac{6.67 \times 10^{-34} \times 3 \times 10^{8}}{2 \times 10^{-7}}$
$-5 \times 1.6 \times 10^{-19}$
$=10.005 \times 10^{-19}-8 \times 10^{-19}$
$=2.005 \times 10^{-19}$
or $V_{0} =\frac{2.005 \times 10^{-19}}{1.6 \times 10^{-19}}$ volt
$=1.25\, V$