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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 38

  • question_answer
    An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom of mass m acquired a result of photon emission will be (R, Rydberg constant and h, Planck's constant)

    A) \[\frac{25\,m}{24\,hR}\]

    B)        \[\frac{24\,m}{25\,hR}\]

    C) \[\frac{150\,hR}{25\,m}\]           

    D)        \[\frac{25\,hR}{24\,m}\]

    Correct Answer: C

    Solution :

    [c] Energy of photon \[=\frac{hc}{\lambda }=hcR\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{5}^{2}}} \right)=\frac{24hcR}{25}\] Momentum of photon \[=\frac{E}{c}=\frac{24hR}{25}\] =Momentum of atom Velocity of atom \[=\frac{24hR}{25m}\]where m = mass of atom.


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