From the Bohr's postulate, the total angular momentum of the moving electron is an integral multiple of h / [2 (pi)]
i.e., mvr = nh / [2 (pi)]
So, 2 (pi) r = nh / mv,
For orbit nearest to the nucleus in hydrogen atom, n = 1,
So, 2 (pi) r = h/(mv) = lamda
i.e. orbit nearest to the nucleus in hydrogen atom accomodates one electron wave.
Radius of the orbit nearest to the nucleus in hydrogen atom =
r1 = [h2 (Epsilon0] / [(pi) m e2]
i.e., r1 = [(6.6 x 10-34)2 x (8.86 x 10-12)] /
[3.14 x (9.1 x 10-31) x (1.6 x 10-19)2]
= 0.528 x 10-10 m.
So, circumference of the orbit nearest to the nucleus in hydrogen atom =
2 x 3.14 x 0.528 x 10-10
= 3.316 x 10-10 m
Now, wavelength of one electron wave =
lamda = h / (m v)
i.e. lamda = h / [square root of (2 m E)]
i.e. lamda = [(6.6 x 10-34 J.s)] / [square root of {2 x (9.1 x 10-31 kg) x
(13.6 x 1.6 x 10-19 J)}]
i.e. lamda = (6.6 x 10-34 J.s) / (19.9 x 10-25) m
i.e., lamda = 3.316 x 10-10 m
i.e. orbit nearest to the nucleus in hydrogen atom accomodates one electron wave.
==============
By :- Dr. A. W. Pangantiwar