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

Options

A. 1000
B. 60
C. 10000
D. 31

Answer

D

Explanation

`(0.001)/(2) xx 10^(3) = (0.5 xx 10^(-16)) n^(2)`
`{(("Because radius of" `n^(th)` "orbit is equal t o")`, `(r_(n) = n^(2)` `r_(0) where r_(0)= 0.529 Å))}`
`:. n^(2) = 1000`
or `n = 31`