Question

The quantum number corresponding to orbit of diameter 0.0001 mm in hydrogen atom will be nearly (Given that the radius of orbit with n = 1 is $0.51\times {10}^{-10}$):

• A. 39
• B. 31
• C. 9
• D. 49

Answer 1

The correct option is D 49

Radius of orbit $=\frac{0.0001}{2}=5\times {10}^{-5}mm=5\times {10}^{-8}m$

Radius of $n^{th}$ orbit is given by :- [from Bohr's model]

$r_n=r_0{n}^2$

where $r_n=$ radius of $1st$ orbit

$=0.51\times {10}^{-10}m$

n: Quantum number of orbit

$\Rightarrow n^2=\frac{r_n}{r_0}=\frac{5\times {20}^{-8}}{0.51\times {10}^{-1}}$

$\Rightarrow n^2\simeq 2381$

So,$n=\sqrt{2381}\simeq 49$