What is the radii of the fourth orbit in the hydrogen atom?

Given h= 6.6238 x 10^-27erg s; e=4.8022 x 10^-10 esu; m= 9.1072 x 10^-28 g; \(\pi\) = 3.1416

8.45 x 10^-8cm

The correct answer is option 1. 8.45 x 10^-8cm.

We know that

\(r_n \text{ = }\frac{n^2h^2}{4\pi^2 me^2}\)

For the 1st orbit of the H- atom,

\(r_1 \text{ = }\frac{1^2h^2}{4\pi^2 me^2}\)

Similarly, for the 4th orbit, the radius will be,

\(r_4 \text{ = }\frac{4^2h^2}{4\pi^2 me^2}\)

Substituting the values  h= 6.6238 x 10^-27erg s; e=4.8022 x 10^-10 esu; m= 9.1072 x 10^-28 g; \(\pi\) = 3.1416, we get

\(r_4 \text{ = }4^2\text{ × }0.529\text{ × }10^{-8}cm\)

\(r_4 \text{ = }8.45{ × }10^{-8}cm\)