Question

When light of wavelength $248 \,nm$ falls on a metal of threshold energy $3.0\, eV$, the de-Broglie wavelength of emitted electrons is_______ $\mathring{A}$.
(Round off to the Nearest Integer).
[Use: $\sqrt{3}=1.73, h =6.63 \times 10^{-34} Js$ $m _{ e }=9.1 \times 10^{-31} kg ;$
$c =3.0 \times 10^{8} ms ^{-1};$
$\left.1 eV =1.6 \times 10^{-19} J \right]$

Answer 1

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

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

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

$=0.05 eV \times 100=5 eV$

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

$5=3+ K.E$

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

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

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

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

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

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

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

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

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

$=9 \mathring{A}$