Question

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.

(b) 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,

${\displaystyle {\text{v}}_1=\frac{{\text{e}}^2}{{\text{n}}_{1}4\pi {\in }_0\left(\frac{\text{h}}{2\pi }\right)}=\frac{{\text{e}}^2}{2{\in }_0\text{h}}}$

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

∴ ${\displaystyle {\text{v}}_1=\frac{{(1.6\times {10}^{-19})}^2}{2\times 8.85\times {10}^{-12}\times 6.62\times {10}^{34}}}$

= 0.0218 × 10⁸

= 2.18 × 10⁶ m/s

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

${\displaystyle {\text{v}}_2=\frac{{\text{e}}^2}{{\text{n}}_{2}2{\in }_0\text{h}}}$

= ${\displaystyle \frac{{(1.6\times {10}^{-19})}^2}{2\times 2\times 8.85\times {10}^{-12}\times 6.62\times {10}^{-34}}}$

= 1.09 × 10⁶ m/s

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

${\displaystyle {\text{v}}_3=\frac{{\text{e}}^2}{{\text{n}}_{3}2{\in }_0\text{h}}}$

= ${\displaystyle \frac{{(1.6\times {10}^{-19})}^2}{3\times 2\times 8.85\times {10}^{-12}\times 6.62\times {10}^{-34}}}$

= 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:

${\displaystyle {\text{T}}_1=\frac{2\pi {\text{r}}_1}{{\text{v}}_1}}$

Where

r₁ = Radius of the orbit

${\displaystyle \frac{{\text{n}}_1^2{\text{h}}^2{\in }_0}{\pi {\text{me}}^2}}$

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

∴ ${\displaystyle {\text{T}}_1=\frac{2\pi {\text{r}}_1}{{\text{v}}_1}}$

= ${\displaystyle \frac{2\pi \times {\left(1\right)}^2\times {(6.62\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{2.18\times {10}^6\times \pi \times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}}$

= 15.27 × 10⁻¹⁷

= 1.527 × 10⁻¹⁶ s

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

${\displaystyle {\text{T}}_2=\frac{2\pi {\text{r}}_2}{{\text{v}}_2}}$

Where,

r₂ = Radius of the electron in n₂ = 2

${\displaystyle =\frac{{\left({\text{n}}_2\right)}^2{\text{h}}^2{\in }_0}{\pi {\text{me}}^2}}$

∴ ${\displaystyle {\text{T}}_2=\frac{2\pi {\text{r}}_2}{{\text{v}}_2}}$

= ${\displaystyle \frac{2\pi \times {\left(2\right)}^2\times {(6.62\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{1.09\times {10}^6\times \pi \times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}}$

= 1.22 × 10⁻¹⁵ s

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

${\displaystyle {\text{T}}_3=\frac{2\pi {\text{r}}_3}{{\text{v}}_3}}$

Where,

r₃ = Radius of the electron in n₃ = 3

= ${\displaystyle \frac{{\left({\text{n}}_3\right)}^2{\text{h}}^2{\in }_0}{\pi {\text{me}}^2}}$

∴ ${\displaystyle {\text{T}}_3=\frac{2\pi {\text{r}}_3}{{\text{v}}_3}}$

= ${\displaystyle \frac{2\pi \times {\left(3\right)}^2\times {(6.62\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{7.27\times {10}^5\times \pi \times 9.1\times {10}^{-31}\times (1.6\times {10}^{-19})}}$

= 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.