Q. Find out the radius of iodine atom. (It is given that for iodine Atomic no. $53$ , Mass no. $126$ )

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A

$2.5\times 10^{- 11}m$

B

$2.5\times 10^{- 9}m$

C

$7\times 10^{- 9}m$

D

$7\times 10^{- 6}m$

Solution:

Radius of atom, $r_{n}$ = $0.529\frac{n^{2}}{Z}\overset{o}{A}$
Atomic number, $Z$ of Iodine $=53$
$n=5$ , Last shell of Iodine having electrons $\left(\right.2$ , $8$ , $18$ , $18$ , $7$ $\left.$
$r_{n}=0.529\frac{\left(\right. 5 \left(\left.\right)^{2}}{53}\overset{o}{A}$
$r_{n}=0.249\overset{o}{A}$
$r_{n}=2.5\times 10^{- 11}m$

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