Question

No of waves made by e^- in Bohr's third orbit in one revolution.?

Answer 1

Number of waves = n(n - 1)/2 where n = Principal quantum number or number of orbit number of waves = 3(3 - 1)/2 = 3 * 2/2 = 3

ALTERNATIVE SOLUTIONS :

In general, the number of waves made by a Bohr electron in an orbit is equal to its quantum number.

According to Bohr’s postulate of angular momentum, in the 3rd orbit

Mur = n h/2π

Mur = 3 (h/2π) …..(i) [n = 3]

According to de Broglie relationship

λ = h/mu ….(ii)

Substituting (ii) in (i), we get

(h/λ) r = 3 (h/2π) or 3λ = 2πr

[∵ mu = h/λ]

Thus the circumference of the 3rd orbit is equal to 3 times the wavelength of electron i.e. the electron makes three revolution around the 3rd orbit.

ALTERNATIVE SOLUTION :

r_n for H = r₁ * n²

r₃ for H = 0.529 * 9 *10^-8 cm

= .529 * 9 *10^-10 m (∵ r₁ = 0.529 Å)

Also u_n = Z u₁/n ;

∴ u₃ = 2.19 *10⁸/3 cm sec^-1 = 2.19 *10+16/3 m sec^-1

(∵ u₁ = 2.19 *10⁸ cm sec^-1)

∴ No. of waves in one round

= 2π r₃/λ = 2π r₃/h/mv₃ = 2π r₃ *m/h

= 2 * 22 *0.529 *9 *10^-10 *2.19 *10⁶ *9.108 * 10^-31/7 *3 *6.62 *10^-34 = 3