Question

In hydrogen atom the kinetic energy of electron is 3.4 eV. The distance of that electron from the nucleus:

• A. 2.116A⁰
• B. 0.529A⁰
• C. 1.587A⁰
• D. 21.16A⁰

Answer 1

The correct option is A

2.116A⁰

K.E=-E

$\to$E=-3.4 eV=$\frac{-13.6}{n^2}$

$\to$$n^2$=$\frac{13.6}{3.4}$=4

n=2

$\to$r=0.529$\times 2^2{A}^o$

=2.116 $A^{\circ }$