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

  • question_answer
    A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be

    A)  \[800\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B)  \[825\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)  \[975\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D)  \[01025\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

    [c] Number of possible spectral lines emitted when an electron jumps back to ground state from nth orbit \[\frac{n\left( n-1 \right)}{2}\] Here,\[\frac{n\left( n-1 \right)}{2}=6\Rightarrow n=4\] Wavelength \[\lambda \] from transition from n = 1 to n = 4 is given by, \[\frac{1}{\lambda }=R\left( \frac{1}{1}-\frac{1}{{{4}^{2}}} \right)\Rightarrow \lambda =\frac{16}{15R}=975\overset{\text{o}}{\mathop{\text{A}}}\,\]


You need to login to perform this action.
You will be redirected in 3 sec spinner