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

  • question_answer
    The ionization potential of H-atom is 13.6 V. When it is excited from ground state by monochromatic radiations of 970.6 A, the number of emission lines will be (according to Bohr's theory)

    A)  10

    B)  8    

    C)  6

    D)  4

    Correct Answer: C

    Solution :

    \[\frac{1}{\lambda }=R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[\Rightarrow \frac{1}{970.6\times {{10}^{-10}}}=1.097\times {{10}^{7}}\left[ \frac{1}{{{1}^{2}}}-\frac{1}{n_{2}^{2}} \right]\Rightarrow {{n}_{2}}=4\] \[\therefore \] Number of emission line \[N=\frac{n\left( n-1 \right)}{2}=\frac{4\times 3}{2}=6\]


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