Calculation of Velocity:

mvr = nh/2π; v = nh/2πmr

By substituting for r we get

v = 2πKZe²/nh

Where except n and Z all are constants

v = 2.18 x 10⁸ Z/n  cm/sec

Further application of Bohr’s work was made, to other one electron species (Hydrogenic ion) such as He⁺ and Li²⁺. In each case of this kind, Bohr’s prediction of the spectrum was correct. 

Illustration:

The velocity of electron in the second orbit of He⁺  will be

(A)   2.18 x 10⁶   m/s               (B)   1.09 x 10⁶ m/s

(C)  4.36 x 10⁶  m/s                (D)    None of these

v,= v_oZ/n. Hence (A) is correct.

Calculate the velocity of an electron in Bohr’s first orbit of hydrogen atom

(Given r = 0.53 x 10^–10 m)

According to Bohr’s theory

mvr = nh/2π or v = nh/2πmr

r = 0.53 x 10^–10m

n = 1, h = 6.62 x 10^–34kg m² s^–1, m = 9.1 x 10^–31 kg

v = 1x (6.62 x 10^-34 kgm²s^-1) / 2 x 22/7 x(9.1x 10^-31kg ) x ( 0.53 x 10^-10m)  = 2.18x 10⁶ m/s

The velocity of e^- in first Bohr’s orbit is 2.17 x 10⁶ m/s. Calculate the velocity in 3^rdorbit of He²⁺ ion.

v_n = v_oZ/n

Hence v = 2.19 x 10⁶x2/3  = 1.45 x 10⁶ m/s