The approximate value of quantum number `n` for the circular orbit of hydrogen of `0.0001 nm` in diameter is

Options

A. 100
B. 60
C. 81
D. 31

Answer

D

Explanation

Diameter of the orbit `=0.0001 mm = 10^(-4)mm`
`" "=10^(-7)m`
`therefore` Radius of the orbit `=(10^(-7)m)/2=0.5 times 10^(-7)m" …(i)"`
The radius of the `n^(th)` circular orbit of the hydrogen atom,
`r_(n)=n^(2)r_(0)=0.53n^(2)`
where `r_(0) approx 0.53 overset(@)(A)=0.53 times 10^(-10)m`
`therefore r_(n)=(0.53 times 10^(-10)n^(2)) " ...(ii)"`
`therefore` From (i) and (ii), we get
`0.5 times 10^(-10)n^(2)=0.5 times 10^(-7)`
`rArr n^(2)=1000" "therefore n approx 31`