JEE Advanced (Fill in the Blanks): Modern Physics | Chapter-wise Tests for JEE Main & Advanced PDF Download

Fill in the Blanks

Q.1. To produce characteristic X-rays using a Tungsten target in an X-ray generator, the accelerating voltage should be greater  than ___________ volts  and the energy of the characteristic radiation is _____eV. 

(The binding energy of the innermost electron in Tungsten is 40 keV).          (1983 - 2 Marks)

Ans. 30,000, 30,000

Solution. For minimum accelerating voltage, the electron should jump from n = 2 to n = 1 level.

For characteristic X-rays

The binding energy of innermost electron = 40 keV 

∴ Ionisation potential of tungsten = 40 kV = 40 × 10³ V

∴ Minimum accelerating voltage,

Q.2. The radioactive decay rate of a  radioactive element is found to be 10³ disintegration/second at a certain time. If the half life of the element is one second , the decay rate after one second is _______ and after three seconds is _______.        (1983 - 2 Marks)

Ans. 500 disintegration/sec, 125 disintegration/sec

Solution.  = Initial activity = 1000 dps (given)

A = Activity after n half lives

Q.3. The maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the ...... of the incident radiation.        (1984- 2 Marks)

Ans. frequency

Solution. Note :  According to law of photoelectric effect 

(K.E.)_max = h_v – hv₀ 

i.e., the maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the frequency of incident radiation. 

Q.4. In the Uranium radioactive series the initial nucleus is  and the final nucleus is When the Uranium nucleus decays to lead, the number of  a-particles emitted is ......and the number of b-particles emitted is .....       (1985 - 2 Marks)

Ans. eight, six

Solution.

First we find the number of a- particles. The change in mass number during the decay from uranium to lead = 238 – 206 = 32. Therefore, the number of a-particles (with mass no. 4) = 32/4 = 8

The change in atomic number (i.e, number of protons) taking place when 8 a-particles are emitted and lead is formed is = 92 – (82 + 2 × 8) = 92 – (82 + 16) = 92 – 98 = – 6

This change will take place by emitting of six β-particles.

Q.5. When the number of electrons striking the anode of an X-ray tube is increased, the ........ of the emitted X-rays increases, while when the speeds of the electrons striking the anode are increased, the cut-off wavelength of the emitted X-rays.........         (1986 - 2 Marks)

Ans. intensity, decreases

Solution. 

Note : More the number of electrons striking the anode, more is the intensity of X-rays.

When the speed of the striking electrons on anode is increased, the emitted X-rays have greater energy. We know that energy, E = hc/λ . Therefore, when E increases then l

decreases.

Q.6. When Boron nucleus  is bombarded by neutrons, a -particles are emitted. The resulting nucleus is of the element ........ and has the mass number ......      (1986 - 2 Marks)
Ans. lithium, 7

Solution.

The resulting nucleus is of element lithium and mass number is 7.

Q.7. Atoms having the same ........... but different ........... are  called isotopes.       (1986 - 2 Marks)

Ans. atomic number, mass number

Solution. Atomic number, mass number

Q.8. The binding energies per nucleon for deuteron (₁H²) and helium (₂He⁴) are 1.1 MeV and 7.0 MeV respectively. The energy released when two deuterons fuse to form a helium nucleus (₂He⁴) is ..........      (1988 - 2 Marks)

Ans. 23.6 MeV

Solution.

Binding energy of two deuterons 

= 2 [1.1 × 2] = 4.4 MeV

Binding energy of helium nucleus = 4 × 7.0 = 28 MeV 

The energy released = 28 – 4.4 = 23.6 MeV

Q.9. In the forward bias arrangement of a p-n junction rectifier, the p end is connected to the ........terminal of the battery and the direction of the current is from ............to ..................in the rectifier.      (1988 - 2 Marks)

Ans. positive, p-part, n-part

Solution. Positive, p-part, n-part

Q.10. ........... biasing of p-n junction offers high resistance to current flow across the junction. The biasing is obtained by connecting the p- side to the .......... terminal of the battery.      (1990 - 2 Marks)

Ans. reverse, negative terminal

Solution. Reverse, negative terminal.

Q.11. The wavelength of the characteristic X-ray K_α line emitted by a hydrogen like element is 0.32 Å. The wavelength of the K_β line emitted by the same element will be ............        (1990 - 2 Marks)

Ans. 0.27Å

Solution. We know that

On dividing, we get λ = 0.27 Å

Q.12. The Bohr radius of the fifth electron of phosphorous atom (atomic number = 15) acting as a dopant in silicon (relative dielectric constant = 12) is .................... Å        (1991 - 1 Mark)

Ans. 3.81Å

Solution. The fifth valence electron of phosphorous is in its third shell, i.e., n = 3. For phosphorous, Z = 15. The Bohr’s radius for nth orbit

Q.13. For the given circuit shown in fig. to act as full wave rectifier, the a.c. input should be connected across .................... and .................... and the d.c. output would appear across .................... and ................       (1991 - 1 Mark)

Ans. B and D, A and C

Solution. B and D is a.c. input and A and C is the d.c. output.

Case (i) When B is –ve and D is +ve Current passes from D → A → C → B 

Case (ii) When B  is + ve and D is – ve 

Current passes from B → A → C → B

Thus curve is always from A to C in output (a d.c. current)

Q.14. In an X-ray tube, electrons accelerated through a potential difference of 15, 000 volts strike a copper target. The speed of the emitted X-ray inside the tube is ................... m/s       (1992 - 1 Mark)

Ans. 3 × 10⁸

Solution. The speed of X-rays is always 3 × 10⁸ m/s in vacuum. It does not depend on the potential differences through which electrons are accelerated in an X-ray tube.

Note : All electromagnetic waves propagate at 3 × 10⁸ m/s in vacuum.

Q.15. In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state n is ....................      (1992 - 1 Mark)

Ans. –1

Solution.

Q.16. In the nuclear process,  ₆C¹¹ → ₅B¹¹ + β⁺+X , X stands for .............      (1992 - 1 Mark)

Ans. neutrino

Solution.  (neutrino)

The balancing of atomic number and mass number is correct.

Therefore, X stands for neutrino.

Q.17. In a ..... biased p-n junction, the net flow of holes is from the n region to the p region.        (1993 - 1 Mark)

Ans. reverse

Q.18. A potential difference of 20 kV is applied across an X-ray tube. The minimum wavelength of X-rays generated is...........Å.     (1996 - 2 Marks)

Ans. 0.62Å

Solution.

Q.19. The wavelength of K_α X-rays produced by an X-ray tube is 0.76Å. The atomic number of the anode material of the tube is.....     (1996 - 2 Marks)

Ans. 41

Solution. 

Q.20. Consider the following reaction :

Mass of the deuterium atom = 2.0141 μ 

Mass of helium atom = 4.0024 μ 

This is a nuclear ............. reaction in which the energy Q released is ............. MeV.        (1996 - 2 Marks)

Ans. fusion, 24.03

Solution. This is a nuclear fusion reaction 

Energy released = (Δm) [931.5 MeV/u] 

= [2 × 2.0141 – 4.0024 ] × 931.5 MeV 

= 24.03 MeV