CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications

(A) Using the Bohr’S Model Calculate the Speed of the Electron in a Hydrogen Atom in the N = 1, 2, and 3 Levels. (B) Calculate the Orbital Period in Each of These Levels. - CBSE (Science) Class 12 - Physics

Login
Create free account


      Forgot password?

Question

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Solution

(a) Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an `electron, νis given by the relation,

`v_1 = e^2/(n_14pi in_0(h/(2pi))) = e^2/(2in_0h)`

Where,

= 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

`:.v_1 = (1.6 xx 10^(-19))^2/(2xx 8.85 xx 10^(-12) xx 6.62 xx 10^34)`

`= 0.0218 xx 10^8 = 2.18 xx 10^(6) "m/s"`

For level n2 = 2, we can write the relation for the corresponding orbital speed as:

`v_2 =  e^2/(n_3 2 in_0 h)`

`= (1.6 xx 10^(-19))/(3xx2xx8.85xx10^(-12) xx 6.62 xx  10^(-34))`

`= 7.27 xx 10^(5) "m/s"`

Hence, the speed of the electron in 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 orbital period of the electron when it is in level n1 = 1.

Orbital period is related to orbital speed as:

`T_1 = (2pir_1)/v_1`

Where

r1 = Radius of the orbit

`(n_1^2 h^2 in_0)/(pime^2)`

h = Planck’s constant = 6.62 × 10−34 Js

e = Charge on an electron = 1.6 × 10−19 C

= Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2

m = Mass of an electron = 9.1 × 10−31 kg

`:. T_1 = (2pir_1)/v_1`

=` (2pi xx (1)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(2.18 xx 10^6 xx pi xx  9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2`

`= 15.27 xx 10^(-17) = 1.527 xx 10^(-16) s`

For level n2 = 2, we can write the period as:

`T_2 = (2pir_2)/v_2`

Where,

r2 = Radius of the electron in n2 = 2

`= ((n_2)^2 h^2 in_0)/(pime^2)`

`:. T_2 = (2pir^2)/v_2`

`= (2pi xx (2)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(1.09 xx 10^6 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2)`

`= 1.22 xx 10^(-15) s`

And, for level n3 = 3, we can write the period as:

`T_3 = (2pir_3)/v_3`

Where,

r3 = Radius of the electron in n3 = 3

`= ((n_3)^2h^2 in_0)/(pime^2)`

`:. T_3 = (2pir_3)/v_3`

`= (2pi xx (3)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(7.27 xx 10^5 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19)))`

`= 4.12 xx 10^(-15) s`

Hence, the orbital period in each of these levels is 1.52 × 10−16 s, 1.22 × 10−15 s, and 4.12 × 10−15 s respectively.

  Is there an error in this question or solution?

APPEARS IN

Solution (A) Using the Bohr’S Model Calculate the Speed of the Electron in a Hydrogen Atom in the N = 1, 2, and 3 Levels. (B) Calculate the Orbital Period in Each of These Levels. Concept: Bohr'S Model for Hydrogen Atom.
S
View in app×