Question

Photoelectrons are liberated by ultraviolet light of wavelength $3000\mathring{A}$ from a metallic surface for which the photoelectric threshold is $4000\mathring{A}$. Calculate de Broglie wavelength of electrons emitted with maximum kinetic energy.

Answer 1

$Explanation:$

From $Einstei{n}^{\prime }sphotoelectricequation$ we get:

$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_{th}}+K.E$

where,

$\lambda =$Wave length$=3000A^o=3000\times {10}^{-8}cm$(given)

${\lambda }_{th}=$threshold wave length$=4000A^o=4000\times {10}^{-8}cm$(given)

$h=$Planck's constant$=6.625\times {10}^{-34}Js$

$c=$Velocity of light$=3\times {10}^{8}m/s$

$K.E=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_{th}}]=6.625\times {10}^{-34}\times 3\times {10}^8[\frac{1}{3000\times {10}^{-8}}-\frac{1}{4000\times {10}^{-8}}]$

$K.E=1.6565\times {10}^{-19}J$

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

$\frac{1}{2}m{v}^2=1.6565\times {10}^{-19}J$

multiplying above equation with $m$(mass of electron)

$m=9.1\times {10}^{-31}kg$

$m^2{v}^2=2\times 1.6565\times {10}^{-19}\times 9.1\times {10}^{-31}$

$mv=5.49\times {10}^{-25}$

de Broglie $\lambda =\frac{h}{mv}=\frac{6.625\times {10}^{-34}}{5.49\times {10}^{-25}}=1.2\times {10}^{-9}=1.2nm$

Hence the correct answer is $\lambda =1.2nm$