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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    The electron in a hydrogen atom makes a transition from \[n={{n}_{1}}\] to \[n={{n}_{2}}\] state. The time period of the electron in the initial state \[({{n}_{1}})\] is eight times that in the final state \[({{n}_{2}})\]. The possible values of \[{{n}_{1}}\] and \[{{n}_{2}}\]are:

    A)  \[{{n}_{1}}=8,{{n}_{2}}=1\]                        

    B)  \[{{n}_{1}}=4,{{n}_{2}}=2\]

    C)  \[{{n}_{1}}=2,{{n}_{2}}=4\]        

    D)  \[{{n}_{1}}=1,{{n}_{2}}=8\]

    Correct Answer: B

    Solution :

    In a hydrogen atom the time period is given by \[T\propto {{n}^{3}}\]                 \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{n}_{1}}}{{{n}_{2}}} \right)}^{3}}\Rightarrow \frac{8}{1}={{\left( \frac{{{n}_{1}}}{{{n}_{2}}} \right)}^{3}}\] \[\therefore \]                  \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{2}{1}\] Thus, \[{{n}_{1}}=4\] and \[{{n}_{2}}=2\]


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