Question

Light of wavelength of $2000\stackrel{o}{A}$ falls on an aluminium surface. In aluminium, 4.2 eV are required to remove an electron from its surface. Calculate the sum of kinetic energy(in electron volt) of the fastest, and the slowest emitted photoelectrons. (Plancks constant $h=6.6\times {10}^{-34}Js$, and speed of light $c=3\times {10}^{8}m{s}^{-1}$)

Answer 1

Energy corresponding to incident photon
$hv=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2000\times {10}^{-10}}$
$=9.9\times {10}^{-19}J$
$=\frac{9.9\times {10}^{-19}}{1.6\times {10}^{-19}}eV=6.2eV$
a. The kinetic energy of fastest electrons,
$E_k=hv-W$
or $E_k=6.2eV-4.2eV=2eV$
b. The kinetic energy of the slowest electron is zero, since the emitted electrons have all possible energies from 0 to certain maximum value $E_k$.
c. If $V_s$ is the stopping potential, then
$E_k=e{V}_s$
or $V_S=\frac{E_k}{e}$
$=\frac{2eV}{e}=2V$
d. If ${\lambda }_0$ is the cut-off wavelength for a aluminium, then
$W=\left(hc/{\lambda }_0\right)$
or ${\lambda }_0=\left(hc/W\right)$
$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4.2\times 1.6\times {10}^{-19}}$
$=3000\times {10}^{-10}m=3000\stackrel{o}{A}$