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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Critical Thinking

  • question_answer
    An electron passing through a potential difference of 4.9 V collides with a mercury atom and transfers it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state                                                                        [AMU (Med.) 2002]

    A)            2050 Å                                     

    B)            2240 Å

    C)            2525 Å                                     

    D)            2935 Å

    Correct Answer: C

    Solution :

               \[\frac{hc}{\lambda }=E\] = eV                    \[\Rightarrow \lambda =\frac{hc}{eV}-\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 4.9}=2525\overset{\circ }{\mathop{A}}\,\]


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