Question

Find the maximum speed v_max of the photoelectrons leaving the surface of silver

under the influence of the ultraviolet radiation of the wave length λ = 0.155μm. The work

function of silver is A = 4.7 eV.

Answer 1

$\lambda=0.155\mu m$

A=4.7eV

We know that

$E=\frac{hc}{\lambda}$

$h=6.625\times10^{-34}Js$

$E=\frac{6.625\times10^{-34}\times3\times10^8}{1550\times1.6\times10^{-19}}$

$E=\frac{12420}{1550}eV$

$E=8eV$

We know that

Photo electric effect

$E-A=eV_s$

$(8-4.7)eV=eV_s$

$eV_s=3.3eV$

We know that

$\frac{1}{2}mv^2=eV_s$

$v=\sqrt\frac{2eV_s}{m}$

$$v=\sqrt{\frac{2\times3.3\times1.6 \times10^{-19}}{13\times9.1\times10^{-31}}}=2.98\times10^{5}m/sec$$