Question

When light of wavelength $248nm$ falls on a metal of threshold energy $3.0eV.$ the de - Broglie wavelength of emitted electrons is $A^{\circ }.$ (Round off to the Nearest Interger).

[Use : $\sqrt{3}=1.73,h=6.63\times {10}^{-34}Js$
$m_e=9.1\times {10}^{-31}kg;c=3.0\times {10}^{8}m{s}^{-1};1eV=1.6\times {10}^{-19}J$]

Answer 1

$\underset{\text{photon}}{\text{Incident energy of}}=\underset{\text{metal}}{\text{Work function}}+\underset{\text{photoelectron}}{\text{K.E. of}}$

$hv=h{v}_0+KE$

$\frac{6.63\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{248\times {10}^{-9}m\times 1.6\times {10}^{-19}Je{V}^{-1}}=3.0+K.E.$

$K.E.=2.0eV$

$\lambda =\frac{h}{\sqrt{2mK.E.}}=\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 2\times {10}^{-19}\times 1.6}}$

$=8.68\times {10}^{-10}m=9A^{\circ }$