Answer:
Option D
Explanation:
Bohr radius (rn) = $\epsilon_{0}n^{2}h^{2}$
$r_{n}=\frac{n^{2}h^{2}}{4\pi^{2}me^{2}kZ}$
$k=\frac{1}{4\pi \epsilon_{0}}$
$r_{n}=\frac{n^{2}h^{2}\epsilon_{0}}{\pi^{}me^{2}Z}$
= $n^{2}\frac{a_{0}}{Z}$
where m = mass of electron
e= charge of electron
h = plank's constant
k= Coulomb constant
$r_{n}= \frac{n^{2}\times 0.53}{Z}\dot{A}$
Radius of nth Bohr orbit for H -atoms
$=0.53 n^{2}\dot{A}$
[ Z= 1 for H-atom]
$\therefore$ Radius of 2nd Bohr orbit for H-atom
=0.53 × (2)2 = $2.12 \dot{A}$
