The energy of electron in an excited hydrogen atom is $- 3.4 e V$ . Its angular momentum according to Bohr's theory will be
$\frac{h}{2 π}$
$\frac{3 h}{2 π}$
$\frac{3}{2 π h}$
$\frac{h}{π}$

Question

The energy of electron in an excited hydrogen atom is $- 3.4 e V$ . Its angular momentum according to Bohr's theory will be
$\frac{h}{2 π}$
$\frac{3 h}{2 π}$
$\frac{3}{2 π h}$
$\frac{h}{π}$

Toppr Admin · Accepted Answer

Given-E-x2212-3-4eVEnergy- E-x2212-13-6eVn2n2-x2212-13-6eV-x2212-3-4eVn2-4n-2From Bohr-apos-s theory- the angular momentum id given byL-nh2-x3C0-L-2-h2-x3C0-L-h-x3C0-The correct option is A-

The energy of electron in an excited hydrogen atom is −3.4eV. Its angular momentum according to Bohr's theory will beh2π3h2π32πhhπ

Given,E=−3.4eVEnergy, E=−13.6eVn2n2=−13.6eV−3.4eVn2=4n=2From Bohr's theory, the angular momentum id given byL=nh2πL=(2)h2πL=hπThe correct option is A.

The energy of electron in an excited hydrogen atom is $- 3.4 e V$ . Its angular momentum according to Bohr's theory will be
$\frac{h}{2 π}$
$\frac{3 h}{2 π}$
$\frac{3}{2 π h}$
$\frac{h}{π}$

Given,
$E = - 3.4 e V$
Energy, $E = \frac{- 13.6 e V}{n^{2}}$
$n^{2} = \frac{- 13.6 e V}{- 3.4 e V}$
$n^{2} = 4$
$n = 2$
From Bohr's theory, the angular momentum id given by
$L = \frac{n h}{2 π}$
$L = \frac{(2) h}{2 π}$
$L = \frac{h}{π}$
The correct option is A.