## Physics Class 12 Term 2 Answer Key

Physics Answer Key 2022 for Class 12 Term 2 Exam: CBSE Physics Class 12 Term 2 Exam has conducted on 20th May 2022 (Friday). No doubt every student after exiting the exam hall will be worried about how many marks he/she would be scoring, which question was wrong and how many questions could be expectedly correct. So instead of waiting for the result, they started searching for the questions asked in the exam. To minimise your fear and anxiety, here in this article, we have shared the complete Class 12 Physics Term 2 Answer Key 2022 at a glance. No need to keep searching here and there just check all the questions with their correct answers.

## Physical Answer Key Class 12 Term 2

Central Board of Secondary Education has divided the Class 12 Physics Board Exam into two Terms. In Term 2 the exam is divided into three sections with 12 questions for which the exam pattern that has been followed is as follows-

1. Section A consists of 1 to 3 questions carrying 2 marks each [Very Short Answer Questions].

2. Section B consists of 4 to 11 questions carrying 3 marks each [Short Answer Questions].

3. Section C consists of 1 question carrying 5 marks [Case Based Question].

## Class 12 Physics Answer Key Term 2

As CBSE will not release the official Class 12 Physics Answer Key for Term 2 Exam along with other subjects, we have provide the unofficial Physics Term 2 Answer Key for Class 12 in this article.

## Class 12 Physics Answer Key Term 2- Questions & Solution

Here, we have provided complete Class 12 Physics Term 2 Answer Key with correct answers one by one. So if are looking for the complete CBSE Class 12 Term 2 Physics Answer Key go through the solutions discussed below.

Question 1- What is meant by the Energy bandgap in a solid? Draw the energy conductor, an insulator and a semiconductor

Solution- Energy Band Gap si the difference in energy between the valence band and the conduction band of solid material (such as an insulator or semiconductor) that consists of the range of energy values forbidden to electrons in the material.

Question 2- Name the device which converts into an ac input signal into a dc output signal. Write the principal of working of the device.

Solution- A rectifier is a device that converts an oscillating two-directional alternating current (AC) into a single-directional direct current (DC). Rectifiers can take a wide variety of physical forms, from vacuum tube diodes and crystal radio receivers to modern silicon-based diode. The rectifier is a device that is capable of converting an alternating current into a unidirectional or pulsating form of direct current. The process of conversion of alternating currents into direct currents is known as rectification.

Question 3- (a) Name the spectral series for a hydrogen atom which lies in the visible region. Find the ratio of the maximum to the minimum wavelengths of this series.

Solution- Balmer series of Hydrogen region lies in visible region of electromagnetic spectrum.

The Balmer series is the set of six named series describing the spectral line emissions of the total hydrogen atoms. The lines are emitted when the electron present in the hydrogen atom transits from n=3 or greater orbital down to n=2 orbital.

According to the equation of Balmer series
1/λ=R(1/n12−1/n22)

=λmin/λmax=(1/22−1/32)/(1/22−1/∞2)

=5/9

Hence, The ratio of minimum to maximum wavelength in the Balmer series is 5: 9.

(b) What are matter waves? A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of the de Broglie wavelength associated with the proton to that with the alpha particle.

Solution- The wave associated with each moving particle is known as a matter wave. The matter-wave has a wavelength of λ=hp, where h is the Planck’s constant and p is the moment of the moving particle.

Using formula λ = h/√2mk, Ratio- 2:1

Question 4- (a) Differentiate between nuclear fission and nuclear fusion

Solution- Fusion is where two light atomic nuclei combine and release energy, while fission is the process of splitting two heavy, unstable atomic nuclei into two lighter nuclei, also releasing energy – although less than with fusion.

1. Nuclear Fusion is the joining of atomic nuclei and Nuclear fission is the splitting of atomic nuclei
2. Nuclear Fusion produces far more energy than that created by Nuclear fission
3. Nuclear Fusion, unlike Nuclear fission, does not create harmful radioactive by-products that need to be stored for thousands of years

(b) Deuterium undergoes fusion as per the reaction :

Find the duration for which an electric bulb of 500 W can be kept glowing by the fusion of 100 g of deuterium.

Solution- 100/2 x 6.023 x 10^-23

Energy Released- 3.27/2 MeV

t = E/P

= 500 x 60 x 60 x 24 x 365

Question 5- In the Geiger-Marsden experiment, calculate the distance of closest approach for an alpha particle with energy 2.56 x 10^–12 J. Consider that the particle approaches the gold nucleus (Z = 79) in head-on position.

Solution- If r0​ be the distance of closest approach between the nucleus and the alpha particle, then the kinetic energy of the alpha particle is Ek​=1/4πϵ0​​ x (Ze)(2e)​/ r0

Thus, r0​∝1/Ek​1

Hence, when the kinetic energy of the alpha particle is doubled, the closet distance r0​ will be halved.

(b) If the above experiment is repeated with a proton of the same energy, then what will be the value of the distance of closest approach?

Question 6- Answer the following, giving reason:

(a) The resistance of a p-n junction is low when it is forward biased and is high when it is reversed biased.

Solution- A small increase in forwarding voltage across p-n junction shows large increase in forwarding current. Hence the resistance (=voitage /current)(=voitage /current) of p-n junction is low when forward biased.
A large increase in reverse voltage across p-n junction shows a small increase in reverse current. Hence the resistance of p-n junction is high when reversing biased.

(b) Doping of intrinsic semiconductors is a necessity for making electronic devices.

Solution- Semiconductors are doped to generate either a surplus or a deficiency in valence electrons. Doping allows researchers to exploit the properties of sets of elements, referred to as dopants, in order to modulate the conductivity of a semiconductor.

(c) Photodiodes are operated in reverse bias.

Solution- The photodiode is operated in reverse biased for operating in the photoconductive mode. As the photodiode is in reverse bias the width of the depletion layer increases. This reduces the junction capacitance and thereby the response time. In effect, the reverse bias causes faster response times for the photodiode.

Question 7- (a) The interference pattern is not observed in Young's double slit experiment when the two sources S, and S, are far apart. Explain

Solution- Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than the wavelength of light is used. Young’s double-slit experiment helped in understanding the wave theory of light which is explained with the help of a diagram. A screen or photodetector is placed at a large distance ’D’ away from the slits as shown. But when we use two independent sources it never act as a coherent source and hence no interference patterns occur at screen and also no fringes are observed as two independent sources are not coherent.

(b) Mention the conditions for the two sources to be coherent.

Solution- The two conditions for two light sources to be coherent are as follows: -

(i) The sources should be monochromatic and originate from a common single source.

(ii) The amplitudes of the waves should be equal.

(c) What is the effect on the interference pattern in a Young's double slit experiment, if the source of wavelength λ is replaced by another source of wavelength 1.5λ, with the interference pattern still observable?

Question 8- A ray of light is incident on a prism at an angle of 45° and passes symmetrically as shown in the figure. Calculate.

Solution-

Physics Class 12 Term 2 Question Paper PDF [Code-54/4/2]