Question

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function $5.01$ eV, when ultraviolet light of $200$ nm falls on it, must be :

• A. $1.2V$
• B. $2.4V$
• C. $-1.2V$
• D. $-2.4V$

Answer 1

The correct option is A $1.2V$

From Einstein's photoelectric equation, $e{V}_0=h\nu -W$

Here, $h\nu =\frac{hc}{\lambda }=\frac{(6.62\times {10}^{-34})\times (3\times {10}^8)}{200\times {10}^{-9}}=9.93\times {10}^{-19}J$ and work function,$W=5.01eV=5.01\times 1.6\times {10}^{-19}J$

Now, $V_0=h\nu /e-W/e=\frac{9.93\times {10}^{-19}}{1.6\times {10}^{-19}}-\frac{5.01\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}}=1.2V$