JIPMER Jipmer Medical Solved Paper-2000

  • question_answer
    The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is \[20,397\text{ }c{{m}^{-1}}\]. The wavelength of energy for the same transition in He+ is :

    A)  \[81,988c{{m}^{-1}}\]                   

    B)  \[40,994c{{m}^{-1}}\]   

    C)         \[20,497c{{m}^{-1}}\]     

    D)         \[5,099\text{ }c{{m}^{-1}}\]

    Correct Answer: D

    Solution :

    As \[\lambda \propto \frac{1}{{{Z}^{2}}}\] Hence wavelength of energy for same transition in He is, here Z = 2 hence,  \[\frac{20397}{4}=5099\text{ }c{{m}^{-1}}\]


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