JEE Main 2016 · 9 Apr · Shift 1 · Physics Q21

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PhysicsSingle correct+4 / -1Question 21 of 90

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 :

\dfrac{hc}e\left[\dfrac1{\lambda_3}-\dfrac1{\lambda_2}-\dfrac1{\lambda_1}\right]

\dfrac{hc}e\left[\dfrac1{\lambda_3}+\dfrac1{\lambda_2}-\dfrac1{\lambda_1}\right]

\dfrac{hc}e\left[\dfrac1{\lambda_3}+\dfrac1{2\lambda_2}-\dfrac3{2\lambda_1}\right]

\dfrac{hc}e\left[\dfrac1{\lambda_3}+\dfrac1{2\lambda_2}-\dfrac1{\lambda_1}\right]

Answer and worked solution

Official answer: C

$eV=\frac{hc}{\lambda_1}-\phi$ and $3eV=\frac{hc}{\lambda_2}-\phi$ give $\phi=\frac{hc}2\left(\frac3{\lambda_1}-\frac1{\lambda_2}\right)$ . Then
$V_3=\frac{hc}e\left[\frac1{\lambda_3}+\frac1{2\lambda_2}-\frac3{2\lambda_1}\right].$

Topics:Photoelectric effect

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JEE Main 2016 · 9 Apr · Shift 1, Physics question 21