Q. What is the radius of Iodine atom? (Atomic no.53, mass no.126)
A) $2.5 \times 10^{- 11}m$
B) $2.5 \times 10^{- 9}m$
C) $7 \times 10^{- 9}m$
D) $7 \times 10^{- 11}m$
Explanation
(a) $\therefore n = 5$
$r_{n} = (0.53 \times 10^{- 10})\frac{n^{2}}{Z}$
$= \frac{0.53 \times 10^{- 10} \times 5^{2}}{53} = 2.5 \times 10^{- 11}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.