Question

Find the wavelength of the electron orbiting in the first excited state in hydrogen atom

Answer 1

According to De Broglie's equation,  λ = h/(mv), where h is the Planck's constant whose value is  6.626 X 10⁻³⁴ Joules and m is the mass of the electron whose value is 9.1 X 10⁻³¹ kg.

The velocity of an electron in the 1st orbit of the hydrogen atom, v’ = 2.18 x 10⁶ m/s

The velocity of an electron in the 2nd orbit (n = 2) of the hydrogen atom, v = v’/n, so, v = 2.18 x 10⁶ / 2 = 1.09 x 10⁶ m/s

So, mv = 9.919 x 10⁻²⁵

According to the equation, λ = (6.626 X 10⁻³⁴) / (9. 919 X 10⁻²⁵) m

λ = 6.680 x 10⁻¹⁰ m

So, the wavelength λ = 6.680 Å

Answer 2

For an electron in first excited state i.e. n = 2

So, if λ be its wavelength (De-Broglie), then we have

n λ = 2πrn 

where rn is the radius of the second orbit.

rn = 0.5 x n2 (in A) = 0.5 x 4 = 2A

2 x λ = 2 x π x 2 A = λ = 2π  (A) = 6.28 A