Important Questions for Class 12th Physics with Answers

Download the list of important questions for class 12th Physics and get a detailed note of each theoretical and numerical question in the form of answers.

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The Important Questions for Class 12th Physics have been prepared as per the NCERT textbooks and the syllabus provided by the CBSE. It is written comprehensively to ensure that the students learn better.

Students studying in class 12th have a lot to do with their studies and are therefore always short of time and having a set of important questions prepared by experts can be a great help for them as they can find more time and make their schedules more elaborate and flexible to cover their syllabus.

This article will pay attention to the questions asked and highlight some of the important questions as an introduction to the students. The questions ranging from multiple choice to short answer-type questions have been provided below, and for the convenience of students, a PDF file with the full notes on important questions has been provided below.

List of NCERT Physics Chapters for Class 12th

Solutions to the following topics in the class 12th physics book are provided in the article. These are the chapters for CBSE class 12th physics.

Alternating Current
Communication Systems
Current Electricity
Electric Charges and Fields
Electromagnetic Induction
Electromagnetic Waves
Electrostatic Potential and Capacitance
Magnetism And Matter
Moving Charges and Magnetism
Ray Optics and Optical Instruments
Semiconductor Electronics Materials Devices and Simple Circuits
Wave Optics

Important Multiple Choice Questions with Answers for Class 12th Physics

A multiple-choice question is an objective assessment where students are asked to select only correct answers from the choices offered as options. Here are some important multiple choice questions as part of class 12th physics questions.

Question 1

What is the angular momentum of an electron in the third orbit of an atom?

(a) 3.15×10−34">3.15×10−343.15×10−34 Js

(b) 3.15×10−30">3.15×10−303.15×10−30 Js

(d) 3.15×10−33">3.15×10−333.15×10−33 Js

Solution: Angular momentum is given by

L = nh2π = 3×6.6×10−34×72×22 = 3.15×10−34"> L = nh2π = 3×6.6×10−34×72×22 = 3.15×10−34L = nh2π = 3×6.6×10−34×72×22 = 3.15×10−34 Js

So (a) is correct

Question 2

Which series of the hydrogen spectrum has wavelengths in the visible range?

(a) Lyman.

(b) Balmer.

(d) Bracket.

Solution: Balmer

Hence (b) is the correct option

Question 3

The value of the fine structure constant is:

(a) 1136">11361136

(b) 1130">11301130

(d) 1137">11371137

Solution: (d)

Question 4

If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation is:

(a) 5R/36

(b) 36/5R

(d) 5/R

Solution: ν¯ = 1λ = R(122−132)">¯ν = 1λ = R(122−132)ν¯ = 1λ = R(122−132)

or λ = 365R">λ = 365Rλ = 365R

(b) is the correct option

Question 5

If an atom moves from 2E energy level to E energy level,the wavelength λ">λλ is emitted. If the transition takes place from 4E/3 energy level to E energy level, the wavelength emitted will be:

(A) λ3">λ3λ3

(B) 3λ">3λ3λ

(D) 4λ3">4λ34λ3

Solution: case -1

2E−E = E = hcλ">2E−E = E = hcλ2E−E = E = hcλ

Case -2

4E3−E = hcλ1">4E3−E = hcλ14E3−E = hcλ1

λ1 = 3λ">λ1 = 3λλ1 = 3λ

Hence (b) is correct

Question 6

Maximum frequency of the emission is obtained for the transition:

(a) n=2 to n=1

(b) n=6 to n=2

(d) n=2 to n=6

Solution: (a) as the energy difference is maximum

Question 7

The radius of an electron orbiting in a hydrogen atom is of the order of:

(a) 10−8">10−810−8 m.

(b) 10−9">10−910−9 m.

(d) 10−13">10−1310−13 m.

Solution: Hence (c) is correct

Question 8

If an electron jumps from the 1st orbit to the 4th orbit then it will:

(a) not lose energy.

(b) absorb energy.

(d) increases and decreases periodically

Solution: (b)

Question 9

The energy of a hydrogen atom in the ground state is -13.6 eV. The energy of a He+ ion in the first excited state will be:

(a)-13.6 eV

(b)-6.8 eV

(c)-54.4.6 eV

(d)-27.2 eV

Solution: for first excited state of He+ ion

n=2, Z=2

E = 13.6Z2n2 = −13.6">E = 13.6Z2n2 = −13.6E = 13.6Z2n2 = −13.6 eV

Hence (a) is the correct choice

Question 10

The energy of an electron in the second orbit of hydrogen atom is E and the energy of electron in 3rd orbit(E3">E3E3) of He+ will be:

(a) E3=16E3">E3=16E3E3=16E3

(b) E3=16E9">E3=16E9E3=16E9

(d) E3=4E3">E3=4E3E3=4E3

Solution: E == E0Z2n2">E == E0Z2n2E == E0Z2n2

For H atom, Z=1, n=2

E == E01222 = E04">E == E01222 = E04E == E01222 = E04

For He+ atom, Z=2, n=3

E3 == E02232 = 4E09">E3 == E02232 = 4E09E3 == E02232 = 4E09

Therefore

E3 = 169E">E3 = 169EE3 = 169E

Hence (b) is correct

Important Short Answer Questions with Answers for Class 12th Physics

Short-answer questions are open-finished inquiries that expect understudies to make an answer. Here are some important short answer-type questions as part of class 12th physics questions.

Question 1

(i) State Bohr's postulates to define stable orbits in hydrogen atoms. How does the De-Broglie hypothesis explain the stability of these orbits?

(ii) A hydrogen atom initially in the ground state absorbs a photon, which excites it to the n=4 level. Estimate the frequency of the photon

Solution: Enth = −13.6n2">Enth = −13.6n2Enth = −13.6n2 eV.

E1 = −13.612 = −13.6">E1 = −13.612 = −13.6E1 = −13.612 = −13.6 eV.

E4 = −13.642 = −0.85">E4 = −13.642 = −0.85E4 = −13.642 = −0.85 eV.

Energy absorbed= -0.85 -(-13.6) = 12.75 eV = 12.75×1.6×10−19 = 20.4×10−19">12.75×1.6×10−19 = 20.4×10−1912.75×1.6×10−19 = 20.4×10−19 J

Frequency is given by

ν = Eh = 20.4×10−196.63×10−34 = 3.08×1015">ν = Eh = 20.4×10−196.63×10−34 = 3.08×1015ν = Eh = 20.4×10−196.63×10−34 = 3.08×1015 Hz

Question 2

Calculate the shortest wavelength of light emitted in the Paschen series of hydrogen spectrum. Which part of the electromagnetic spectrum does it belong? Explain the above observations on the basis of Einstein's photoelectric equation.

Solution: Paschen series of hydrogen spectrum is given

ν¯ = 1.1×107(132−1n2)m−1">¯ν = 1.1×107(132−1n2)m−1ν¯ = 1.1×107(132−1n2)m−1

where n = 4,5,6

1λ = 1.1×107(132−1n2)m−1">1λ = 1.1×107(132−1n2)m−11λ = 1.1×107(132−1n2)m−1

For Shortest wavelength, n = ∞">n = ∞n = ∞

1λ = 1.1×107(132−1∞2)">1λ = 1.1×107(132−1∞2)1λ = 1.1×107(132−1∞2)

λ = 8199Ao">λ = 8199Aoλ = 8199Ao

The series lies in the infrared region of the EM spectrum

Note:

Although we have physics all around us, it is pertinent to understand it theoretically and numerically, which can be difficult. Therefore, based on this rationale, each chapter has the maximum chance to be asked in the term I examination. Also, make use of the PDF given above.