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) = 0.5xx10^(-7) m`
The radius of the nth circular orbit of the hydrogen atom
`r_n = n^2r_0 = 0.5 n^2 " where " r_0 = 0.5 Å = 0.5xx10^(-10) m`
`therefore r_n = (0.5xx10^(-10))n^2 " "....(2)`
`therefore " From (1) and (2) ", 0.5xx10^(-10) n^2 = 0.5xx10^(-7)`
`therefore n^2 = 1000 therefore n = 31`