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

Answer

2

Explanation

Velocity of electron `= 1/(275) xx ` velocity of light
`=1/(275) xx 3.0xx10^(10) = 1.09 xx 10^8` cm/sec
Now `V_n = (2pi e^2)/(nh) implies n = (2pi e^2)/(V_n xx h)`
`=(2xx 3.14 xx (4.803xx 10^(-10))^2)/(1.09xx10^8 xx 6.625 xx 10^(-27)) = 2.006 = 2`