JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    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   JEE Main Online Paper (Held On 09 April 2016)

    A) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}+\frac{1}{2{{\lambda }_{2}}}-\frac{3}{2{{\lambda }_{1}}} \right]\]

    B) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}+\frac{1}{{{\lambda }_{2}}}-\frac{3}{{{\lambda }_{1}}} \right]\]

    C) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}+\frac{1}{2{{\lambda }_{2}}}-\frac{1}{{{\lambda }_{1}}} \right]\]               

    D) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}-\frac{1}{{{\lambda }_{2}}}-\frac{1}{{{\lambda }_{1}}} \right]\]

    Correct Answer: A

    Solution :

                    \[eV=\frac{hc}{{{\lambda }_{1}}}-{{\phi }_{0}}\]                                \[3eV=\frac{hc}{{{\lambda }_{2}}}-{{\phi }_{0}}\]                 \[eV'=\frac{hc}{{{\lambda }_{3}}}-{{\phi }_{0}}\] using these equations \[V'=\frac{hc}{e}\left( \frac{1}{{{\lambda }_{3}}}+\frac{1}{2{{\lambda }_{2}}}-\frac{3}{2{{\lambda }_{1}}} \right)\]


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