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JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    Electrons in hydrogen like atom (Z=3) make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Find the stopping potential for the photoelectrons ejected by the longer wavelength, then. (Rydberg constant)

    A)  5V

    B)  2V  

    C)  0.754 V

    D)  2.99V

    Correct Answer: C

    Solution :

    [c] For transition from n=5 to n=4, \[hv=13.6\times 9\left[ \frac{1}{16}-\frac{1}{25} \right]=\frac{13.6\times 9\times 9}{16\times 25}=2.754eV\]For transition from n=4 to n=3, \[hv'=13.6\times 9\left[ \frac{1}{9}-\frac{1}{16} \right]=\frac{13.6\times 9\times 7}{9\times 16}=5.95eV\] For transition n=4 to n=3, the frequency is high and hence wavelength is short. For photoelectric effect, \[hv'=-W=e{{v}_{0}}\] Where W= work function \[5.95\times 1.6\times {{10}^{-19}}-W=1.6\times {{10}^{-19}}\times 3.95\] \[\Rightarrow W=2\times 1.6\times {{10}^{-19}}=2eV\] Again applying \[hv-W=eV{{'}_{0}}\] We get, \[2.754\times 1.6\times {{10}^{-19}}-2\times 1.6\times {{10}^{-19}}=\] \[1.6\times {{10}^{-19}}V{{'}_{0}}\]


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