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Question: An electron in Bohr’s hydrogen atom has an energy of –3.4 eV. The angular momentum of the electron i...

An electron in Bohr’s hydrogen atom has an energy of –3.4 eV. The angular momentum of the electron is

A

h/π

B

h/2π

C

nh/2π (n is an integer)

D

2h/π

Answer

h/π

Explanation

Solution

The energy of an electron in an orbit of principal quantum number n is given as

E = −13.6n2 6mueV\frac{- 13.6}{n^{2}}\mspace{6mu} eV

⇒ -3.4 eV = − 6mu13.6n2 6mueV\frac{- \mspace{6mu} 13.6}{n^{2}}\mspace{6mu} eV

⇒ n2 = 4 ⇒ n = 2

The angular momentum of an electron in nth orbit is given as

L = nh2π 6mu\frac{nh}{2\pi}\mspace{6mu}, Putting n = 2

We obtain L = 2h2π 6mu 6mu= 6muhπ\frac{2h}{2\pi}\mspace{6mu}\mspace{6mu} = \mspace{6mu}\frac{h}{\pi}