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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    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)             [JEE Main 8-4-2019 Afternoon]

    A) \[\frac{1}{2}\lambda \]                                  

    B) \[\frac{3}{4}\lambda \]

    C) \[\frac{2}{3}\lambda \]                      

    D)   \[\frac{4}{9}\lambda \]

    Correct Answer: D

    Solution :

    \[hv-\phi =KE\]           \[\Rightarrow {{\left( \frac{hc}{\lambda } \right)}_{incident}}=KE+\phi \]           \[{{\left( \frac{hc}{\lambda } \right)}_{incident}}\simeq KE\]           \[KE=\frac{{{p}^{2}}}{2m}=\frac{hc}{{{\lambda }_{incident}}}=\frac{hc}{\lambda }\]                             ?(1) \[\Rightarrow \]\[\frac{{{p}^{2}}\times {{(1.5)}^{2}}}{2m}=\frac{hc}{\lambda '}\]                            ...(2) divide and \[{{(1.5)}^{2}}=\frac{\lambda }{\lambda '}\]\[\Rightarrow \lambda '=\frac{4\lambda }{9}\]


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