Question

Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of the hydrogen atom. The ground state energy of the hydrogen atom is −13.6 eV.

Answer 1

Given: Ground state energy = −13.6 eV

For a hydrogen atom,

E_n = `(-13.6)/n^2 eV`

First excited state ⇒ n = 2

E₂ = `(-13.6)/2^2`

= `(-13.6)/4`

= −3.4 eV

In the Bohr model:

K.E. = ∣E_n​∣ = 3.4 eV

= 3.4 × 1.6 × 10⁻¹⁹

= 5.44 × 10⁻¹⁹ J

By using de-Broglie relation:

λ = `h/p`

Since,

K.E.= `p^2/(2 m)`

p = `sqrt(2 m K.E.)`

Thus,

λ = `h/sqrt(2 m K.E.)`

= `(6.63 xx 10^-34)/(sqrt(2 xx 9.1 xx 10^-31 xx 5.44 xx 10^-19))`

= 6.66 × 10⁻¹⁰ m

= 6.66 Å