Question

No of waves in 3rd orbit of H is

Answer 1

hey mate here is your answer..

No.of waves = circumference of electron orbit / wave length. =2πr /λ λ=h /mv. No.of waves = 2π/h *(mvr) 

Answer 2

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 :

rn for H = r1 * n2

r3 for H = 0.529 * 9 *10-8 cm

= .529 * 9 *10-10 m (∵ r1 = 0.529 Å)

Also un = 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

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