2. Solved Papers
  3. BCECE Medical

BCECE Medical BCECE Medical Solved Papers-2010

  • question_answer
    An electron jumps from the first excited state to the ground state of hydrogen atom. What 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%


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