Question

The work function of caesium is 2.14 eV. Find a) the threshold frequency for caesium and b) the wavelength of the incident light if photocurrent is brought to zero by stopping potential of 0.60 V.

Answer 1

a) $W=hv$

Where $W$ = work function

$h$ = planck constant

$v$ = threshold frequency

$v=\frac{W}{h}=\frac{2.14\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}=5.2\times {10}^{14}hertz$

b) Wavelength of incident photon =$\frac{hc}{ev}+W$

By substituting the values we get

Wavelength = $4.5\times {10}^{-7}m$