Question

The work function for caseium 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 $500nm$. Calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer 1

work function for caseium atom $=1.9ev$

$\left(1\right)$ threshold wavelength.

$\frac{hc}{{\lambda }_0}=1.9ev$

${\lambda }_0=\frac{hc}{1.9ev}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1.9\times 1.6\times {10}^{-19}}m$

$=6.5388\times {10}^{-7}m$

$=653.88nm$

$2)$ threshold frequency

${\lambda }_0=\frac{c}{\lambda }=\frac{3\times {10}^8}{6.5388\times {10}^{-7}}$

$4.58799\times {10}^{14}$

$=4.6\times {10}^{14}Hz$

Now caseium is irradiated with a wavelength of $500nm$

$EnergyE=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{500\times {10}^{-9}}$

$3.97\times {10}^{-19}$

$E=K.E+{\varphi }_0$

$K.E=E-{\varphi }_0$

$=3.97\times {10}^{-19}-1.9\times 1.6\times {10}^{-19}$

$=0.93\times {10}^{-19}$ joule $=0.58ev$

$\frac{1}{2}m{v}^2=K.E$

$V=\sqrt{\frac{2K.E}{m}}$

$=\sqrt{\frac{2\times 0.93\times {10}^{-19}}{9.1\times {10}^{-31}}}$

$\sqrt{0.20439\times {10}^{12}}$

$=0.4521\times {10}^6$

$=4.521\times {10}^{5}m/s$