Question

The work function for the caesium atom is $1.9eV$. Calculate

(a) the threshold wavelength and

(b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength of $500nm$, calculate the kinetic energy and the velocity of the photoelectron.

Answer 1

The work function of cesium atom is 1.9 eV. Converting the unit in J.
$1.9eV\times 1.602\times {10}^{-19}J/eV=3.04\times {10}^{-19}J$

The threshold frequency is $\frac{3.04\times {10}^{-19}}{6.626\times {10}^{-34}}=4.59\times {10}^{14}$ Hz.

The threshold wavelength is ${\lambda }_0=\frac{c}{{\nu }_0}=\frac{3.0\times {10}^8}{4.59\times {10}^{14}}=6.54\times {10}^{-7}$ m.

The energy of radiated light is
$E=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{500\times {10}^{-9}}=3.98\times {10}^{-19}J$

The kinetic energy of ejected electron is
$K.E=3.98\times {10}^{-19}-3.04\times {10}^{-19}=9.4\times {10}^{-20}J=\frac{1}{2}m{v}^2$
$v=\sqrt{\frac{2K.E}{m}}=\sqrt{2\frac{9.14\times {10}^{-20}}{9.1\times {10}^{-31}}}=4.54\times {10}^{5}m/s$