A) \[\frac{L}{2}\]
B) \[\frac{L}{4}\]
C) 4L
D) 2L
Correct Answer: B
Solution :
From the formula for momentum is given by \[p=\sqrt{2m{{E}_{k}}}\] So, \[mr\omega =\sqrt{2m{{E}_{k}}}\](As \[p=m\upsilon =mr\omega \]) or \[mr(2\pi n)=\sqrt{2m{{E}_{k}}}\] \[(\because \omega =2\pi n)\] or \[rn\propto \sqrt{{{E}_{k}}}\] Hence, \[\frac{{{r}_{1}}{{n}_{1}}}{{{r}_{2}}{{n}_{2}}}=\frac{\sqrt{{{E}_{{{k}_{1}}}}}}{{{E}_{{{k}_{2}}}}}\] \[\left( \text{Given,}\,\frac{{{E}_{{{k}_{1}}}}}{{{E}_{{{k}_{2}}}}}=2\,\text{and}\,\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{1}{2} \right)\] Putting the given values, we get \[\frac{{{r}_{1}}}{{{r}_{2}}}\times \frac{1}{2}=\sqrt{2}\] or \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{1}{2\sqrt{2}}\] Now, angular momentum \[L=m\upsilon r\] or \[L=m{{r}^{2}}\omega =m\times {{r}^{2}}\times 2\pi n\]or \[L\propto {{r}^{2}}\omega \] or \[\frac{{{L}_{2}}}{{{L}_{1}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{1}{8}\times 2=\frac{1}{4}\] \[{{L}_{2}}=\frac{{{L}_{1}}}{4}=\frac{L}{4}\]
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