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

Question

The energy of electron in an excited hydrogen atom is -3-4 eV- Its angular momentum according to Bohr-s theory will be-cfrac-h-pi -cfrac-h-2pi -cfrac-3h-2pi -cfrac-2h-pi -

Toppr Admin · Accepted Answer

E-x2212-13-6n2eV-x2212-13-6n2-xA0-3-4-x21D2-n-2So- its angular momentum -nh2-x3C0-xA0-2h2-x3C0-xA0-h-x3C0-

The energy of electron in an excited hydrogen atom is -3.4 eV. Its angular momentum according to Bohr's theory will be:hπh2π3h2π2hπ

E=−13.6n2eV−13.6n2= 3.4⇒n=2So, its angular momentum =nh2π =2h2π =hπ

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

$E = \frac{- 13.6}{n^{2}} e V$
$\frac{- 13.6}{n^{2}} = 3.4$
$\Rightarrow n = 2$
So, its angular momentum $= \frac{n h}{2 π}$ $= \frac{2 h}{2 π}$ $= \frac{h}{π}$