Q102 / 120 · VITEEE 2013 · Physics

As the electron in Bohr's orbit of hydrogen atom passes from state $n=2$ to, $n=1$, the $\mathrm{KE}(\mathrm{K})$ and the potential energy (U) changes as

  • Correct answer
    A
    $K$ four fold, $U$ also four fold
  • B
    $K$ two fold, $V$ also two fold
  • C
    $K$ four fold, $U$ two fold
  • D
    $K$ two fold, $U$ four fold

Solution

$\mathrm{KE}$ of an electron in nth orbit : $\mathrm{K}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^{2}}$ and PE of an electron in nth orbit : $\mathrm{U}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^{2}}$ $\therefore$ When an electron passes from state $\mathrm{n}=2$ to $\mathrm{n}=1$ $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{1^{2}}{2^{2}}=\frac{1}{4}$ $\begin{array}{l} \text { or } \mathrm{K}_{1}=4 \mathrm{~K}_{2} \\ \frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{l}^{2}}{2^{2}}=\frac{1}{4} \\ \text { or } \mathrm{U}_{1}=4 \mathrm{U}_{2} \end{array}$

Hence, the correct answer is option: 'A'.

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