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JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    A hydrogen like atom (atomic number \[Z\]) is in a higher excited state a quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies \[10.2\,\,eV\] and \[16.8\,\,eV\] respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies \[4.25\,\,eV\] and \[5.95\,\,eV\] respectively. The values of n and \[Z\] are respectively (Ground state energy of hydrogen atom is \[-13.6\,\,eV\])

    A)  6 and 6                                  

    B)  3 and 3

    C)  6 and 3         

    D)  3 and 6

    Correct Answer: C

    Solution :

    \[{{E}_{n}}=\frac{1.36{{Z}^{2}}}{{{n}^{2}}}eV\] If n = 6 and Z = 3 then\[{{E}_{2}}=-\,30.4eV,\]\[{{E}_{3}}=-\,13.6\,eV,\]\[{{E}_{4}}=-\,7.65\,eV,\]\[{{E}_{5}}=-\,4.9\,eV\] and \[{{E}_{6}}=-\,3.4\,eV\] \[\Rightarrow {{E}_{6}}-{{E}_{3}}=10.2eV\]and \[{{E}_{3}}-{{E}_{2}}=16.8\,eV\] Also \[{{E}_{6}}-{{E}_{4}}=4.25\,eV\] and \[{{E}_{4}}-{{E}_{3}}=6.95\,eV\] \[\therefore n=6\]and\[Z=3\]


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