A) \[0.03\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[0.12\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[2.0\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[8.0\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
The radius of Bohr or bit, \[r\propto {{n}^{2}}\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{n}_{1}}}{{{n}_{2}}} \right)}^{2}}\] \[\Rightarrow \] \[{{r}_{2}}={{r}_{1}}{{\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)}^{2}}\] ?(1) Given:\[{{r}_{1}}=0.5\,\overset{{}^\circ }{\mathop{\text{A}}}\,\],\[{{n}_{1}}=1,{{n}_{2}}=4\] Putting given values in eq. (1) \[\therefore \] \[{{r}_{2}}=0.5{{\left( \frac{4}{1} \right)}^{2}}\] \[\Rightarrow \] \[{{r}_{2}}=0.5\times 16\] \[\therefore \] \[{{r}_{2}}=8\,\overset{{}^\circ }{\mathop{\text{A}}}\,\]
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