Question

When light of wavelength $248nm$ falls on a metal of threshold energy $3.0eV$, the de-Broglie wavelength of emitted electrons is_______ $\mathring{A}$.
(Round off to the Nearest Integer).
[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

Answer: 9

Energy incident $=\frac{hc}{\lambda }$

$=\frac{6.63\times 10^{-34}\times 3.0\times 10^8}{248\times 10^{-9}\times 1.6\times 10^{-19}}eV$

$=\frac{6.63\times 3\times 100}{248\times 1.6}$

$=0.05eV\times 100=5eV$

Now using $E=\varphi +K.E$

$5=3+K.E$

$K.E.=2eV=3.2\times 10^{-19}J$

for debroglie wavelength $\lambda =\frac{h}{mv}$

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

so $v=\sqrt{\frac{2KE}{m}}$

hence $\lambda =\frac{h}{\sqrt{2KE\times m}}$

$=\frac{6.63\times 10^{-34}}{\sqrt{2\times 3.2\times 10^{-19}\times 9.1\times 10^{-31}}}$

$=\frac{6.63}{7.6}\times \frac{10^{-34}}{10^{-25}}=\frac{66.3\times 10^{-10}m}{7.6}$

$=8.72\times 10^{-10}m$

$\approx 9\times 10^{-10}m$

$=9\mathring{A}$