Question

The angular momentum of an electron in a Bohr's orbit of H-atom is $4.2178\times {10}^{-34}Kg{m}^2/sec$. Then electron belongs to

• A. n = 1
• B. n = 2
• C. n = 3
• D. n = 4

Answer 1

The correct option is D

n = 4

$n\frac{h}{2\pi }$=$n\frac{6.625\times {10}^{-34}}{2\times 3.14}$=$4.217\times {10}^{-34}$

n=$\frac{4.214\times 2\times 3.14}{6.625}$

n=4