First slide

Bohr Model of the Hydrogen atom.

Question

Angular momentum of the electron in a hydrogen atom is $\frac{3 h}{2 π}$ . Then energy of the electron is

Easy

A

-3.4 eV
B

-10.4 eV
C

-2.36 eV
D

-1.511 eV

Solution

$L_{n} = n . \frac{h}{2 π} \Rightarrow \frac{3 h}{2 π} = n \frac{h}{2 π} \Rightarrow n = 3$
$E_{3} = - \frac{13.6}{3^{2}} e V = - 1.511 e V$

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