Question

The de broglie wavelength of an electron in the 3rd orbit of hydrogen atom is
(answer in angstrom)

Answer 1

Radius of 3rd orbit = r = R(3)^2
= 9 R
[Here R = Radius of first orbit of hydrogen atom]

mvr = nh / (2 π)
h/(mv) = 2πr / n
λ = h / (mv) = 2π × 9R / 3
λ = 6πR
λ = 6 π × (0.529 Å)
λ = 9.97 Å

De-Broglie wavelength is 9.97 Å

Answer 2

Answer:

Explanation:

radius of He 3rd orbit is

0.529 x n²/z A

where n=3 and z=2

0.529 x 9/2

=2.3805 A

we know

mvR = nh/2pi

h/mv=2piR/n---------------1

wavelength = h/mv---------2

equating 1 & 2

wavelength=2piR/n

=2 x 22 x 2.3805

        3 x 7

=4.98 A