Question

An electron in hydrogen atom stays in its second orbit for 10⁻⁸ s. How many revolutions will it make around the nucleus at that time?

An electron in hydrogen atom stays in its second orbit for 10⁻⁸ s. How many revolutions will it make around the nucleus at that time?

[Given: e = 1.6 × 10^-19 C, m = 9.1 × 10^-31 kg]

Answer 1

Bohr Model Background

The time period T_n​ for an electron in the nnn-th orbit is given by:

`T_n = (2pir_n)/v_n`

Radius of n-th orbit:

r_n ​= n² ⋅ r₁ ​where r₁​ = 5.29 × 10⁻¹¹ m

Speed in n-th orbit:

`v_n = v_1/n  "where"  v_1 = 2.18 xx 10^6` m/s

So for n = 2:

r²​ = 4 ⋅ 5.29 × 10⁻¹¹ = 2.116 × 10⁻¹⁰ m

`v^2 = (2.18xx10^6)/2 = 1.09 xx 10^6` m/s

`T_2 = (2pir_2)/v_2 = (2pixx2.116 xx 10^-10)/(1.09 xx 10^6)`

`T_2 = (13.29 xx 10^-10)/(1.09 xx 10^6)` = 1.219×10⁻¹⁵s

t = 10⁻⁸s

Number of revolutions `= t/T^2 = 10^-8/1.219xx10^-15`

= 8.2 × 10⁶ revolutions