Question

One milliwatt of light of wavelength $4560\mathring{A}$ is incident on a caesium surface. Calculate the electron current liberated. Assume a quantum efficiency of $0.5\%$

Answer 1

E = h$\nu$

= $\frac{hc}{\lambda }$ = $\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4560\times {10}^{-10}J}$

= $4.34\times {10}^{-19}J$

1mW of light energy is equivalent to $\frac{{10}^{-3}}{(4.34\times {10}^{-19})}$

= $2.3\times {10}^{15}$ phtons/s

The quantum efficiency is 0.5%

The means that only 0.5% of these photons release photoelectrons.

so, the number of electrons released form the surface per second

= $2.3\times {10}^{15}\times 0.5/100$

= $1.15\times {10}^{13}$ electrons/s

The electron current = $1.15\times {10}^{13}\times 1.6\times {10}^{-19}A$

= $1.84\mu A$