```Calculation of Velocity:
We know that
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
Solution:
v,= voZ/n. Hence (A) is correct.

Illustration:
Calculate the velocity of an electron in Bohr’s first orbit of hydrogen atom
(Given r = 0.53 x 10–10 m)
Solution:
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

Illustration:
The velocity of e- in first Bohr’s orbit is 2.17 x 106 m/s. Calculate the velocity in 3rdorbit of He2+ ion.
Solution:
vn = voZ/n
Hence v = 2.19 x 106x2/3  = 1.45 x 106 m/s```
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