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

(A) \frac{{hc}}{e}\left[ {\frac{1}{{{\lambda _3}}} + \frac{1}{{{\lambda _2}}} - \frac{1}{{{\lambda _1}}}} \right]

(B) \frac{{hc}}{e}\left[ {\frac{1}{{{\lambda _3}}} + \frac{1}{{{2\lambda _2}}} - \frac{1}{{{\lambda _1}}}} \right]

(C) \frac{{hc}}{e}\left[ {\frac{1}{{{\lambda _3}}} - \frac{1}{{{\lambda _2}}} - \frac{1}{{{\lambda _1}}}} \right]

JEE Mains · Physics · STD 12 - Dual nature of radiation and matter

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

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Verified Solution

Correct Answer

(D) \(\frac{{hc}}{e}\left[ {\frac{1}{{{\lambda _3}}} + \frac{1}{{{2\lambda _2}}} - \frac{3}{{{2\lambda _1}}}} \right]\)

Detailed explanation

From Einstein's photoelectric equation, we have

\(\frac{h c}{\lambda_{1}}=\frac{h c}{\lambda_{0}}+e V\) ..... \((1)\)

\(\frac{\mathrm{hc}}{\lambda_{2}}=\frac{\mathrm{hc}}{\lambda_{0}}+3 \mathrm{eV}\) ..... \((2)\)

\(\frac{\mathrm{hc}}{\lambda_{3}}=\frac{\mathrm{hc}}{\lambda_{0}}+\mathrm{eV'}\) ..... \((3)\)

Fromequation \((1)\) and \((2)\)

\(\frac{3}{2 \lambda_{1}}-\frac{2}{2 \lambda_{2}}=\frac{1}{\lambda_{0}}\)

\(\frac{{{\text{hc}}}}{{{\lambda _3}}} - {\text{hc}}\left[ {\frac{3}{{2{\lambda _1}}} - \frac{1}{{2{\lambda _2}}}} \right] = {\text{eV'}}\)

\(\frac{{hc}}{e}\left[ {\frac{1}{{{\lambda _3}}} - \frac{3}{{2{\lambda _1}}} + \frac{1}{{2{\lambda _2}}}} \right] = {V^\prime }\)

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