(a) Using the Bohr’s model calcula | Class 12 Physics Chapter Atoms, Atoms NCERT Solutions

Question:

(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.

Answer:

(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, ν1 is given by the relation,

ν 1 = e2/n14πϵ0(h/2π) = e2/2ϵ0h

Where, e = 1.6 × 10−19 C

ϵ0 = Permittivity of free space = 8.85 × 10-12 N−1 C2 m−2

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

∴ ν1 = (1.6x10-19)2/2x8.85x10-12x6.62x10-34 = 0.0218 x 108 = 2.18 x 106 m/s

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

ν2 = e2/n20h = (1.6x10-19)2/2x2x8.85x10-12x6.62x10-34 = 1.09 x 106 m/s

And, for n3 = 3, we can write the relation for the corresponding orbital speed as:

ν3 = e2/n30h = (1.6x10-19)2/3x2x8.85x10-12x6.62x10-34 = 7.27 x 105 m/s

Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 10 6 m/s, 1.09 × 10 6 m/s, 7.27 × 10 5 m/s respectively.

 

(b) Let T 1 be the orbital period of the electron when it is in level n1 = 1.

Orbital period is related to orbital speed as:

T1 = 2πr1 1

Where, r1 = Radius of the orbit

= n12h2ϵ0/πme2

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

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

ϵ0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2

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

∴ T1 = 2πr11

= (2πx(1)2x(6.62x10-34)2x8.85x10-12)/2.18x106xπx9.1x10-31x(1.6x10-19)2

= 15.27x10-17 = 1.527x10-16 s

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

T2 = 2πr22

Where, r2 = Radius of the electron in n2 = 2

= (n2)2h2ϵ0/πme2

∴ T2 = 2πr22

= (2πx(2)2x(6.62x10-34)2x8.85x10-12)/1.09 x 106 x π x 9.1 x 10-31 x (1.6 x 10-19)2

= 1.22 x 10-15 s

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

T3 = 2πr33

Where, r 3 = Radius of the electron in n 3 = 3

= (n3)2h2ϵ0/πme2

∴ T3 = 2πr33

= (2πx(3)2x(6.62x10-34)2x8.85x10-12)/7.27 x 105 x π x 9.1 x 10-31 x (1.6 x 10-19)2

= 4.12 x 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.

 


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Atoms.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

Comments

Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 7: (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1....