When light of wavelength 248 nm falls on
Question:

When light of wavelength $248 \mathrm{~nm}$ falls on a metal of threshold energy $3.0 \mathrm{eV}$, the de-Broglie wavelength of emitted electrons is ____________\AA

(Round off to the Nearest Integer).

$\left[\right.$ Use: $\sqrt{3}=1.73, \mathrm{~h}=6.63 \times 10^{-34} \mathrm{~J}_{\mathrm{S}}$$\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg} ; \mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}$ $\left.\mathrm{leV}=1.6 \times 10^{-19} \mathrm{~J}\right]$

Solution:

(9)

Energy $=\frac{h c}{\lambda}$

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

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

$=0.05 \mathrm{eV} \times 100=5 \mathrm{eV}$

Now using $\mathrm{E}=\phi+\mathrm{K} . \mathrm{E}$

$5=3+\mathrm{K} . \mathrm{E}$

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

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

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

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

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

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

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

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

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

= 9 \ AA

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