JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

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 ________ $\overset{o}{A}$ . (Round off to the Nearest Integer).

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

Official previous-year question

Held on 16 Mar 2021 · Verified 6 Jul 2026.

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Solution

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

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

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

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

Now using

$E=\phi +K.E$

$5=3+K.E.$

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

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

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

so $v=\sqrt{\frac{2K.E.}{m}}$

hence $\lambda =\frac{h}{\sqrt{2K.E.\times m}}$

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

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

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

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

$=9\overset{o}{A}$

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JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

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

Official previous-year question

Held on 16 Mar 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

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

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

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

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

Now using

$E=\phi +K.E$

$5=3+K.E.$

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

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

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

so $v=\sqrt{\frac{2K.E.}{m}}$

hence $\lambda =\frac{h}{\sqrt{2K.E.\times m}}$

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

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

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

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

$=9\overset{o}{A}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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JEE Main 2021 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2021 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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