JEE Main 2016 — Physics Modern Physics

2016mcqhard

When photons of wavelength ${\lambda }_{1}$ are incident on an isolated sphere, the corresponding stopping potential is found to be $V$ . When photons of wavelength ${\lambda }_{2}$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength ${\lambda }_{3}$ is used then find the stopping potential for this case:

Official previous-year question

Held on 9 Apr 2016 · Verified 6 Jul 2026.

Options

A
$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}+\frac{1}{{\lambda }_{2}}-\frac{1}{{\lambda }_{1}}]$
B
$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}+\frac{1}{2{\lambda }_{2}}-\frac{1}{{\lambda }_{1}}]$
C
$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}-\frac{1}{{\lambda }_{2}}-\frac{1}{{\lambda }_{1}}]$
D
$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}+\frac{1}{{2\lambda }_{2}}-\frac{3}{{2\lambda }_{1}}]$

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Solution

$\frac{hc}{{\lambda }_{1}}=\frac{hc}{{\lambda }_{0}}+eV$ ...(i)

$\frac{hc}{{\lambda }_{2}}=\frac{hc}{{\lambda }_{0}}+3eV$ ...(ii)

$\frac{hc}{{\lambda }_{3}}=\frac{hc}{{\lambda }_{0}}+e{V}^{'}$ ....(iii)

Equation (i) and (ii)

$\frac{3}{2{\lambda }_{1}}-\frac{1}{2{\lambda }_{2}}=\frac{1}{{\lambda }_{0}}$

$\frac{hc}{{\lambda }_{3}}-hc[\frac{3}{2{\lambda }_{1}}-\frac{1}{2{\lambda }_{2}}]=e{V}^{'}$

$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}-\frac{3}{2{\lambda }_{1}}+\frac{1}{2{\lambda }_{2}}]={V}^{'}$

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Options

$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}+\frac{1}{{\lambda }_{2}}-\frac{1}{{\lambda }_{1}}]$

$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}+\frac{1}{2{\lambda }_{2}}-\frac{1}{{\lambda }_{1}}]$

$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}-\frac{1}{{\lambda }_{2}}-\frac{1}{{\lambda }_{1}}]$

$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}+\frac{1}{{2\lambda }_{2}}-\frac{3}{{2\lambda }_{1}}]$

Solution

$\frac{hc}{{\lambda }_{1}}=\frac{hc}{{\lambda }_{0}}+eV$ ...(i)

$\frac{hc}{{\lambda }_{2}}=\frac{hc}{{\lambda }_{0}}+3eV$ ...(ii)

$\frac{hc}{{\lambda }_{3}}=\frac{hc}{{\lambda }_{0}}+e{V}^{'}$ ....(iii)

Equation (i) and (ii)

$\frac{3}{2{\lambda }_{1}}-\frac{1}{2{\lambda }_{2}}=\frac{1}{{\lambda }_{0}}$

$\frac{hc}{{\lambda }_{3}}-hc[\frac{3}{2{\lambda }_{1}}-\frac{1}{2{\lambda }_{2}}]=e{V}^{'}$

$\frac{hc}{e}[\frac{1}{{\lambda }_{3}}-\frac{3}{2{\lambda }_{1}}+\frac{1}{2{\lambda }_{2}}]={V}^{'}$

JEE Main 2016 — Physics Modern Physics

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JEE Main 2016 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics