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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    In an inelastic collision an electron excites a hydrogen atom from its ground state to a M-shell state. A second electron collides instantaneously with the excited hydrogen atom in the M-shell state and ionizes it. At least how much energy the second electron transfers to the atom in the M-shell state?

    A)  +3.4eV                                

    B)  +1.51eV

    C)  -3.4 eV                                

    D)  -1.51 eV

    Correct Answer: B

    Solution :

                     The ground state of hydrogen \[(n=1)\] is represented by K, the first excited state \[(n=2)\] is represented by L, the second excited state \[(n=3)\] is represented by M. So, the energy transferred by the second electron can be given by \[E=-13.6\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[=-13.6\left( \frac{1}{{{3}^{2}}}-\frac{1}{^{{{\infty }^{2}}}} \right)=-1.51\,\text{eV}\]


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