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

  • question_answer
    The energy of electron in the nth orbit of  hydrogen atom is expressed as \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV.\] The shortest and longest wavelength of Lyman series will be

    A)  \[910\overset{\text{o}}{\mathop{\text{A}}}\,,\,\,1213\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B)         \[5463\overset{\text{o}}{\mathop{\text{A}}}\,,\,\,7858\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)  \[1315\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,,\text{ }1530\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,~~\]

    D)         None of these

    Correct Answer: A

    Solution :

    [a] Not Available \[\frac{1}{{{\lambda }_{\max }}}=R\left[ \frac{1}{{{\left( 1 \right)}^{2}}}-\frac{1}{{{\left( 2 \right)}^{2}}} \right]\Rightarrow {{\lambda }_{\max }}=\frac{4}{3R}\approx 1213\overset{\text{o}}{\mathop{\text{A}}}\,\] and \[\frac{1}{{{\lambda }_{\min }}}=R\left[ \frac{1}{{{\left( 1 \right)}^{2}}}-\frac{1}{\infty } \right]\Rightarrow {{\lambda }_{\min }}=\frac{1}{R}\approx 910\overset{\text{o}}{\mathop{\text{A}}}\,\]


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