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BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    An electron Jumps from the first excited state to the ground State of hydrogen atom. Whit will be the percentage change in the speed of electron?

    A)  25%                                      

    B)  50%

    C)  100%                   

    D)         200%

    Correct Answer: B

    Solution :

    Velocity of electron in nth shell \[{{v}_{n}}\propto \frac{1}{n}\]                 So,          \[\frac{{{v}_{2}}}{{{v}_{1}}}=\frac{1}{2},\]                 ie,           \[{{v}_{2}}=\frac{{{v}_{1}}}{2}\] \[\Delta v={{v}_{1}}-{{v}_{2}}=\frac{{{v}_{1}}}{2}\] Hence, % change \[=50%\]


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