The velocity of an electron in a certain Bohr's orbibit of H-atom bears the ratio ` 1: 275 ` to the velocity of light . The find the quantum number ( n) of orbit .

Answer

2

Explanation

Velocity of elctron `= 1/( 275) xx "Velocity of light"`
` = 1/( 275) xx 3,. 0 xx 10^(10) = 1. 089 xx 10^8 cm//sec`
Now `u_n = (2 pi e)/( nh)` ltbRgt ` n= (2 pe^2)/( u_n xx h)`
` = ( 2 xx 3. 14 xx ( 4. 803 xx 10^(10))^2)/( 1.09 xx 10^8 xx 6.625 xx 10^(-27)) =2 . 00 6 ~~2`.