Question

Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of $3.3V$. If photo electric emission begins in this metal at a frequency of $8\times {10}^{14}Hz$, calculate the work function (in eV) for this metal.

Answer 1

Given that;

Stoping potential, $v_s=3.3$v

Minimum frequency needed to eject electrons from metal$=$ frequency for work function $={\nu }_{Th}=$threshold frequency

${\nu }_{Th}=8\times {10}^{14}Hz$

$\therefore \varphi =$ work function $=h{\nu }_{Th}=6.62\times {10}^{-34}\times 8\times {10}^{14}$

$\varphi =5.296\times {10}^{-19}J$

$\because 1.6\times {10}^{-19}J=1eV$

$\varphi =\frac{5.296\times {10}^{-19}}{1.6\times {10}^{-19}}=3.31eV$

Work function for the metal $=\varphi =3.31$eV

Use Einstein equation:-

Incident energy, $E=\varphi +e{v}_s$

$\Rightarrow E=(3.31+3.3)eV$

$E=6.61ev=(6.61\times {10}^{-19}\times 1.6)J$

$E=h\nu =6.61\times 1.6\times {10}^{-19}$

$\Rightarrow \nu =\frac{6.61\times 1.6\times {10}^{-19}}{6.62\times {10}^{-34}}=1.6\times {10}^{15}Hz$

Thus, frequency of light $=\nu =1.6\times {10}^{15}Hz$.