A) \[2.5\times {{10}^{-11}}m\]
B) \[2.5\times {{10}^{-9}}m\]
C) \[7\times {{10}^{-9}}m\]
D) \[7\times {{10}^{-6}}m\]
Correct Answer: A
Solution :
Electronic configuration of iodine is 2, 8, 18, 18, 7, Here \[{{r}_{n}}=(0.053\times {{10}^{-9}}m)\frac{{{n}^{2}}}{Z}\] Here \[n=5\]and \[_{Z-1}{{X}^{A-4}}{{\xrightarrow{_{0}{{\gamma }^{0}}}}_{Z-1}}{{X}^{A-4}}\]hence \[{{r}_{n}}=2.5\times {{10}^{-11}}m\].
You need to login to perform this action.
You will be redirected in
3 sec
