A) \[21.16\times {{10}^{-11}}m\]
B) \[15.87\times {{10}^{-11}}m\]
C) \[10.58\times {{10}^{-11}}m\]
D) \[2.64\times {{10}^{-11}}m\]
Correct Answer: A
Solution :
Radius of Bohr orbit, \[r\propto {{n}^{2}}\] \[\Rightarrow \] \[\frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{n}_{1}}}{{{n}_{2}}} \right)}^{2}}\] \[\therefore \] \[\frac{5.29\times {{10}^{-11}}}{{{r}_{2}}}={{\left( \frac{1}{2} \right)}^{2}}\] \[\Rightarrow \] \[{{r}_{2}}=5.29\times {{10}^{-11}}\times 4\] \[=21.16\times {{10}^{-11}}m\]You need to login to perform this action.
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