Question

An electron in Bohr model of hydrogen atom makes a transition from energy level −1.51 eV to −3.40 eV. Calculate the change in the radius of its orbit. The radius of orbit of electron in its ground state is 0.53 A.

Answer 1

Energy of an electron:

ε_n = `(-13.6)/n^2` eV

For ε_i:

−1.51 = `(-13.6)/n_i^2`

⇒ `n_i^2 = (-13.6)/-1.51`

≈ 9

For ε_f:

 −3.40 = `(-13.6)/n_f^2`

⇒ `n_f^2 = (-13.6)/(-3.40)`

≈ 4

Bohr radius (r_n) = n²r₁

So, r_i = `n_i^2r_1`

= 9 × 0.53

= 4.77 Å

r_f = `n_f^2r_1`

= 4 × 0.53

= 2.12 Å

Change in radius (Δr) = 4.77 − 2.12

= 2.65 Å