NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

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

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

    C)        \[\frac{25\,m}{24\,h\,R}\]           

    D)        \[\frac{24\,m}{25\,h\,R}\] (m is the mass of the electron, R Rydberg constant and h Planck's constant)

    Correct Answer: A

    Solution :

    Here, \[{{E}_{5}}-{{E}_{1}}=\frac{hc}{\lambda }\]and\[\frac{R\,hc}{25}-R\,hc=\frac{hc}{\lambda }\] \[\frac{24}{25}R\,=\frac{1}{\lambda }\]But\[p=\frac{h}{\lambda }\]and\[v=\frac{h}{m\lambda }\]                                 \[=\frac{24}{25}\frac{Rh}{m}\]


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