Question

1. Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
2. Calculate the orbital period in each of these levels.

Answer 1

(a) Let v₁ be the orbital speed of the electron in a hydrogen atom in the ground state level, n₁ = 1.

For charge (e) of an electron, v₁ is given by the relation,

`v_1 = e^2/(n_1 4pi in_0(h/(2pi))) = e^2/(2in_0h)`

Where,

e = 1.6 × 10⁻¹⁹ C

∈₀ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C² m⁻²

h = Planck’s constant = 6.62 × 10⁻³⁴ Js

∴ `v_1 = (1.6 xx 10^(-19))^2/(2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^-34)`

= 0.0218 × 10⁸

= 2.18 × 10⁶ m/s

For level n₂ = 2, we can write the relation for the corresponding orbital speed as:

`v_2 = e^2/(n_2 2 in_0 h)`

= `(1.6 xx 10^(-19))^2/(2 xx 2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^(-34))`

= 1.09 × 10⁶ m/s

And, for n₃ = 3, we can write the relation for the corresponding orbital speed as:

`v_3 = e^2/(n_3 2 in_0 h)`

= `(1.6 xx 10^(-19))^2/(3 xx 2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^(-34))`

= 7.27 × 10⁵ m/s

Hence, the speed of the electron in a hydrogen atom in n = 1, n = 2 and n = 3 is 2.18 × 10⁶ m/s, 1.09 × 10⁶ m/s, 7.27 × 10⁵ m/s respectively.

(b) Let T₁ be the orbital period of the electron when it is in level n₁ = 1.

Orbital period is related to orbital speed as:

`T_1 = (2pir_1)/v_1`

Where,

r₁ = Radius of the orbit = `(n_1^2 h^2 in_0)/(pime^2)`

h = Planck’s constant = 6.62 × 10⁻³⁴ Js

e = Charge on an electron = 1.6 × 10⁻¹⁹ C

∈₀ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C² m⁻²

m = Mass of an electron = 9.1 × 10⁻³¹ kg

∴ `T_1 = (2pir_1)/v_1`

= `(2pi xx (1)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(2.18 xx 10^6 xx pi xx  9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2`

= 15.27 × 10⁻¹⁷

= 1.527 × 10⁻¹⁶ s

For level n₂ = 2, we can write the period as:

`T_2 = (2pir_2)/v_2`

Where, r₂ = Radius of the electron in n₂ = 2

`= ((n_2)^2 h^2 in_0)/(pime^2)`

∴ `T_2 = (2pir_2)/v_2`

= `(2pi xx (2)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(1.09 xx 10^6 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2)`

= 1.22 × 10⁻¹⁵ s

And, for level n₃ = 3, we can write the period as:

`T_3 = (2pir_3)/v_3`

Where,

r₃ = Radius of the electron in n₃ = 3

= `((n_3)^2h^2 in_0)/(pime^2)`

∴ `T_3 = (2pir_3)/v_3`

= `(2pi xx (3)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(7.27 xx 10^5 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19)))`

= 4.12 × 10⁻¹⁵ s

Hence, the orbital period in each of these levels is 1.52 × 10⁻¹⁶ s, 1.22 × 10⁻¹⁵ s and 4.12 × 10⁻¹⁵ s, respectively.