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Class XII – Physics – 1 – MS

Session Ending Exam, 2021-22

Class-XII

Subject –Physics (042)

Max. Marks 35                                                                                                          Max. Time 2 Hrs

General Instructions:

(i) There are 12 questions in all. All questions are compulsory.

(ii) This question paper has three sections: Section A, Section B and Section C.

(iii) Section A contains three questions of two marks each,

Section B contains eight questions of three marks each,

Section C contains one case study-based question of five marks.

(iv) There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the choices in such questions.

(v) You may use log tables if necessary but use of calculator is not allowed.

SECTION – A

1. The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~ mA). What is the reason, then, to operate the photodiode in reverse bias?

Consider the case of an n-type semiconductor. The majority carrier density (n) is considerably larger than the minority hole density p (i.e., n >> p). On illumination, let the excess electrons and holes generated be Dn and Dp, respectively: n′ = n + Δn ; p′ = p + Δp

Here n′ and p′ are the electron and hole concentrations at any particular illumination and n and p are carrier concentration when there is no illumination. Remember Δn = Δp and n > > p. Hence, the fractional change in the majority carriers (i.e., Δn/n) would be much less than that in the minority carriers (i.e., Δp/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity

2. Calculate the orbital period of the electron in the first excited state of hydrogen atom.

OR

Define ionization energy. How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge?

The minimum energy, required to free the electron from the ground state of the hydrogen atom, is known as ionization energy of that atom.

so when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron, ionization energy increases by 200 times.

3. The circuit shown in the figure has two oppositely connected ideal diodes connected in parallel. Find the current flowing through each diode in the circuit.

Diode D1 is reverse biased, so it offers an infinite resistance. So no current flows in the branch of diode D1. Diode D2 is forward biased, and offers negligible resistance in the circuit. So current in the branch

SECTION – B

4. Draw the circuit diagram of a half wave rectifier and explain its working. Also, give the input and output waveforms.

Half wave rectifier:

It consists of a diode D connected in series with load resistor RL across the secondary windings of a step-down transformer. Primary of transformer is connected to a.c. supply. During positive half cycle of input a.c., end A of the secondary winding becomes positive and end B negative. Thus, diode D becomes forward biased and conducts the current through it. So, current in the circuit flows from A to B through load resistor RL.

During negative half cycle of input a.c., end A of the secondary winding becomes negative and end B positive. Thus, diode D becomes reverse biased and does not conduct any current. So, no current flows in the circuit. Since electric current through load RL flows only during positive half cycle, in one direction only i.e., from A to B, so d.c. is obtained across RL.

5. The following graph shows the variation of photocurrent for a photosensitive metal:

(a) Identify the variable X on the horizontal axis.

The variable X on the horizontal axis is collector plate potential.

(b) What does the point A on the horizontal axis represent?

The point A on the horizontal axis represents stopping potential.

(c) Draw this graph for three different values of frequencies of incident radiation υ1, υ2 and υ3 (υ1 > υ2 > υ3) for same intensity.

(d) Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I1 > I2 > I3) having same frequency.

6. Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels.

According to Bohr’s postulates, in a hydrogen atom, as single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit of a given radius, the centripetal force is provided by coulomb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small.

7. (a) How is the size of a nucleus experimentally determined? Write the relation between the radius and mass number of the nucleus. Show that the density of nucleus is independent of its mass number.

Nucleus was first discovered in 1911 by Lord Rutherford and his associates by experiments on scattering of α-particle by atoms. He found that the scattering result could be explained, if atoms consists of a small, central, massive and positive core surrounded by orbiting electron. The experiment results indicated that the size of the nucleus is of the order of 10–14 metres and thus 10,000 times smaller than the size of atom.

Relation between the radius and mass number of the nucleus R = R0 A1/3

If m is the average mass of a nucleon and R is the nuclear radius, then mass of nucleus = mA, where A is the mass number of the element.

This shows that the nuclear density is independent of A.

(b) The nuclear radius of  27Al13 is 3.6 fermi. Find the nuclear radius of  64Cu29.

8. Two wavelengths of sodium light 590 nm and 596 nm are used, in turn to study the diffraction taking place at a single slit of aperture 2 × 10–4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

Given that: Wavelength of the light beam,

ƛ1 = 590 nm = 5.9 ×10–7 m

Wavelength of another light beam,

ƛ2 = 596 nm = 5.96 × 10–7 m

Distance of the slits from the screen = D = 1.5 m

Slits width = a = 2 × 10–4 m

For the first secondary maxima,

9. A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, person has to use what type and power of the correcting lens?

OR

In Young’s double experiment, a monochromatic light of wavelength 5400 Å produces a fringe width of 3 mm. If this source is replaced by another source of monochromatic light of wavelength 6300 Å, then find the fringe width.

10. An object placed at a distance of 16 cm from a convex lens produces an image of magnification m(m > 1). If the object is moved towards the lens by 8 cm then again an image of magnification m is obtained. What is the numerical value of the focal length of the lens?

11. A double convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm.

OR

In a single slit diffraction experiment, the width of the slit is reduced to half its original width. How would this affect the size and intensity of the central maximum?

(a) If width of slit is reduced to half then the size of central maxima will become double.

(b) If width of slit is reduced to half its original width then the intensity of central maximum will be one-fourth.

SECTION – C

12. CASE STUDY : SPEED OF AN ELECTROMAGNETIC WAVE

(iv) (c): b-rays consists of electrons which are not electromagnetic in nature.

(v) (b): The velocity of electromagnetic waves in free space (vacuum) is equal to velocity of light in vacuum (i.e., 3 × 108 m s–1).