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NEET AIPMT SOLVED PAPER MAINS 2010

  • question_answer
    The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in nth state \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV)\]

    A)  5.1 V                    

    B)  12.1 V  

    C)  17.2 V                  

    D)  7 V

    Correct Answer: D

    Solution :

    For \[n=1,{{E}_{1}}=-\frac{13.6}{{{(1)}^{2}}}=-13.6\,eV\] and for \[n=3,{{E}_{3}}=-\frac{13.6}{{{(3)}^{2}}}=-1.51\,eV\] So, required energy \[E={{E}_{3}}-{{E}_{1}}=-(1.51)-(-13.6)\]                 \[=12.09\,eV\] \[\because \]     \[E=W+eV\] \[\therefore \]  \[eV=E-W\]                 \[eV=(12.09-5.1)e\]                 \[V=7\,volt\]


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