Question

The threshold wavelength of silver is $3800\mathring{A}$. Calculate the maximum kinetic energy in eV of photoelectron emitted, when ultraviolet light of wavelength $2600\mathring{A}$ falls on it. (Planck's constant, $h=6.63\times {10}^{-34}J.s.$, Velocity of light in air, $c=3\times {10}^{8}m/s$)

Answer 1

Given : ${\lambda }_0=3800\mathring{A}=3.80\times {10}^{-7}m$

$\lambda =2600\mathring{A}=2.60\times {10}^{-7}m$

To find : ${K.E.}_{max}$,

${K.E.}_{max}=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]$

${K.E.}_{max}=6.63\times {10}^{-34}\times 3\times {10}^8\times [\frac{1}{3.80\times {10}^{-7}}-\frac{1}{2.60\times {10}^{-7}}]$

$=19.89\times {10}^{-26}\times [2.60\times {10}^7-3.80\times {10}^7]$

$=19.89\times {10}^{-26}\times (-0.121\times {10}^7)$

$=-\frac{2.41\times {10}^{-20}}{1.67\times {10}^{-19}}=-1.44eV$

${K.E.}_{max}=1.44eV$