Question -
Answer -
(a) Let ν1 be the orbital speed of the electron in ahydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation,
Where,
e = 1.6 × 10−19 C
∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
h = Planck’s constant = 6.62 × 10−34 Js
For level n2 = 2, we can write the relation for thecorresponding orbital speed as:
And, for n3 = 3, we can write the relation for thecorresponding orbital speed as:
Hence, the speed of the electronin a hydrogen atom in n = 1,n=2, and n=3 is 2.18 × 106 m/s,1.09 × 106 m/s, 7.27 × 105 m/s respectively.
(b) Let T1 be the orbitalperiod of the electron when it is in level n1 = 1.
Orbitalperiod is related to orbital speed as:
Where,
r1 =Radius of the orbit
h =Planck’s constant = 6.62 × 10−34 Js
e = Charge on an electron = 1.6 × 10−19 C
∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
m = Mass of an electron = 9.1 × 10−31 kg
For level n2 = 2, we can write the period as:
Where,
r2 =Radius of the electron in n2 = 2
And, for level n3 = 3, we can write the period as:
Where,
r3 =Radius of the electron in n3 = 3
Hence, the orbital period in eachof these levels is 1.52 × 10−16 s,1.22 × 10−15 s, and 4.12 × 10−15 s respectively.