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# (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

#### 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.

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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.
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