Question

Show that radius (r_n) of the n^th Bohr orbit varies directly with square of the principal quantum number (n) of the orbit.

Answer 1

Let r be the orbit’s radius, and e, m, and v represent the electron's charge, mass, and velocity, respectively. Z is the atomic number, and the nucleus has a positive charge of +Ze.

Centripetal force is provided by the electrostatic force of attraction.

Thus,

`(mv^2)/r = 1/(4 pi epsilon_0)`

or mv² = `(Ze^2)/(4 pi epsilon_0 r)`    ...(1)

From Bohr’s quantisation condition, we have oi

mvr = `n h/(2 pi)`    ...(2)

n = 1, 2, 3, .... is the principal quantum number.

Squaring equation (2) and dividing by equation (1), we get

`(m^2v^2r^2)/(mv^2) = (n^2h^2//4 pi^2)/(Ze^2//4 pi epsilon_0 r)`

⇒ r = `(n^2h^2epsilon_0)/(pi m Ze^2)`

Since h, ε₀, m, Z, and e are constants for an element.

Thus, r_n ∝ n₂