JEE Main 2019 — Chemistry Physical Chemistry

2019mcqmedium

If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda ,$ then for $1.5 p$ momentum of the photoelectron, the wavelength of the light should be:

(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

A
$\frac{4}{9}\lambda$
B
$\frac{3}{4}\lambda$
C
$\frac{2}{3}\lambda$
D
$\frac{1}{2}\lambda$

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Solution

$h\nu -\phi =KE$

${(\frac{hc}{\lambda })}_{incident}=KE+\phi$

Given $(\phi <<KE)$

${(\frac{hc}{\lambda })}_{incident}\approx KE$

${(\frac{hc}{\lambda })}_{incident}=\frac{{p}^{2}}{2m}$ ......(i)

$(\frac{hc}{\lambda '})=\frac{{(1.5p)}^{2}}{2m}$ …….(ii)

Divide (i) and (ii)

$\frac{{\lambda }^{'}}{\lambda }=\frac{{p}^{2}}{2.25{p}^{2}}$

${\lambda }^{'}=\frac{\lambda \times 100}{2.25}=\frac{4}{9}$

${\lambda }^{'}=\frac{4}{9}\lambda$

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Solution

$h\nu -\phi =KE$

${(\frac{hc}{\lambda })}_{incident}=KE+\phi$

Given $(\phi <<KE)$

${(\frac{hc}{\lambda })}_{incident}\approx KE$

${(\frac{hc}{\lambda })}_{incident}=\frac{{p}^{2}}{2m}$ ......(i)

$(\frac{hc}{\lambda '})=\frac{{(1.5p)}^{2}}{2m}$ …….(ii)

Divide (i) and (ii)

$\frac{{\lambda }^{'}}{\lambda }=\frac{{p}^{2}}{2.25{p}^{2}}$

${\lambda }^{'}=\frac{\lambda \times 100}{2.25}=\frac{4}{9}$

${\lambda }^{'}=\frac{4}{9}\lambda$

JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

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

Keep practicing Physical Chemistry