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
Answer & explanation
Correct answer: option 1
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
\(r_4 \text{ = }4^2\text{ × }0.529\text{ × }10^{-8}cm\)
\(r_4 \text{ = }8.45{ × }10^{-8}cm\)