Check out and download important physics questions for class 12th with answers in PDF file format and assess your preparation level.
Class 12th physics important questions are a significant asset for students getting ready for the class 12th board examination. Here you can get the class 12th important questions for physics-based on the NCERT textbook for class 12th. Material science class 12th important questions come in handy to score good grades on board exams. Here we have covered important questions on all points for class 12th physics subject.
Table of Contents
- NCERT physics chapters for class 12th
- Important multiple choice questions with answers for class 12th physics
- Important short answer questions with answers for class 12th physics
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.
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
(c) 3.15×10−31">3.15×10−313.15×10−31 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
Which series of the hydrogen spectrum has wavelengths in the visible range?
Hence (b) is the correct option
The value of the fine structure constant is:
If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation is:
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
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:
Solution: case -1
2E−E = E = hcλ">2E−E = E = hcλ2E−E = E = hcλ
4E3−E = hcλ1">4E3−E = hcλ14E3−E = hcλ1
λ1 = 3λ">λ1 = 3λλ1 = 3λ
Hence (b) is correct
Maximum frequency of the emission is obtained for the transition:
(a) n=2 to n=1
(b) n=6 to n=2
(c) n=1 to n=2
(d) n=2 to n=6
Solution: (a) as the energy difference is maximum
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.
(c) 10−11">10−1110−11 m.
(d) 10−13">10−1310−13 m.
Solution: Hence (c) is correct
If an electron jumps from the 1st orbit to the 4th orbit then it will:
(a) not lose energy.
(b) absorb energy.
(c) release energy.
(d) increases and decreases periodically
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:
Solution: for first excited state of He+ ion
E = 13.6Z2n2 = −13.6">E = 13.6Z2n2 = −13.6E = 13.6Z2n2 = −13.6 eV
Hence (a) is the correct choice
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:
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
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.
(i) State Bohr's postulates to define stable orbits in hydrogen atoms. How does 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
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 infrared region of the EM spectrum
Although we have physics all around us, it is crucial to understand it theoretically and numerically can be difficult. Based on that, each chapter has the maximum chances to occur in the upcoming exam. Also make use of the PDF given, which will make your studies easier. In this article, you can find important questions for class 12th physics. You can download the important questions free of cost in PDF file format. These questions will help you with your class 12th exam.