IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions on Nuclear Physics

Our greatest weakness lies in giving up. The most certain way to succed is always to try just one more time.

– Thomas Alva Edison

The following three questions [(i), (ii) and (iii)] from nuclear physics were included under Linked Comprehension Type Multiple Choice Questions (single answer type) in the IIT-JEE 2009 question paper. They are simple as you will realize.

Paragraph for Questions (i), (ii) and (iii)

Scientists are working hard to develop unclear fusion reactor. Nuclei of heavy hydrogen, ₁H², known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D–D reaction is ₁H²+₁H² → ₂He³ + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ₁H² nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t₀ before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt₀ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 10¹⁴ s/cm³.

It may be helpful to use the following : Boltzmann constant k = 8.6 × 10^–5 eV/K; e²/4πε₀ = 1.44×10^–9 eVm

Question (i):

In the core of nuclear fusion reactor, the gas becomes plasma because of

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between the deuteron–electron pairs

(D) the high temperature maintained inside the reactor core.

The correct option is (D) since the plasma state is achieved at high temperatures.

Question (ii):

Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10^–15 m is in the range

(A) 1.0×10⁹ K < T < 2×10⁹ K

(B) 2.0×10⁹ K < T < 3×10⁹ K

(C) 3.0×10⁹ K < T < 4×10⁹ K

(D) 4.0×10⁹ K < T < 5×10⁹ K

The total kinetic energy of the two deuterons is 2×1.5 kT = 3 kT. At the required separation r (equal to 4×10^–15 m), the entire kinetic energy gets converted into electrostatic potential energy:

3 kT = e²/4πε₀r

Or, 3×8.6×10^–5×T = 1.44×10^–9/(4×10^–15) since e²/4πε₀ = 1.44×10^–9 eVm

Therefore, T = 1.4×10⁹ K. [Option (A)].

Question (iii):

Results of calculations for four different designs of a fusion reactor using D–D reaction are given below. Which of these is most promising based on Lawson criterion ?

(A) deuteron density = 2.0×10¹² cm^–3, confinement time = 5.0×10^–3 s

(B) deuteron density = 8.0×10¹⁴ cm^–3, confinement time = 9.0×10^–1 s

(C) deuteron density = 4.0×10²³ cm^–3, confinement time = 1.0×10^–11 s

(D) deuteron density = 1.0×10²⁴ cm^–3, confinement time = 4.0×10^–12 s.

Lawson number is nt₀. Out of given options, Lawson number is greater than 5 × 10¹⁴ s/cm³ for option (B) since n = 8.0×10¹⁴ cm^–3 and t₀ = 9.0×10^–1 s.

Therefore the correct option is (B).

You will find some useful multiple choice questions on nuclear physics here

AIEEE 2008- An Imaginary Question on Bohr Model

The following question was included in the AIEEE 2008 question paper:

Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr’s model to this system, the radius of the n^th orbital of the system is found to be ‘r_n’ and the kinetic energy of the electron to be ‘T_n’. Then which of the following is true?

(1) T_n α 1/n², r_n α n²

(2) T_n independent of n, r_n α n

(3) T_n α 1/n, r_n α n

(4) T_n α 1/n, r_n α n²

The force k/r supplies the centripetal force for the circular motion of the electron so that we have

k/r = mv²/r where ‘m’ is the mass and ‘v’ is the speed of the electron.

Therefore, mv² = k which is independent of the quantum number ‘n’. The kinetic energy T_n of the electron is ½ mv² which is therefore independent of the quantum number ‘n’.

Also, v = √(k/m).

The angular momentum of the electron in the n^th orbit of radius r_n is mvr_n and in the Bohr model mvr_n = nh/2π where ‘h’ is Planck’s constant. Substituting for ‘v’ we have

m×√(k/m) ×r_n = nh/2π

This gives r_n α n. So the correct option is (2).

In the above question the attractive force on the electron was imagined to be inversely proportional to the distance just for the sake of testing your problem solving skill. In a real hydrogen atom the force is certainly inversely proportional to the square of the distance. You will find questions on real Bohr model on this site by clicking on the label ‘Bohr model ‘ below this post.

All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 (AIPMT 2009) Questions on Nuclear Physics

Everything should be made as simple as possible, but not simpler

– Albert Einstein

The following questions on nuclear physics appeared in the All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 question paper:

1. In the nuclear decay given below:

^AX_Z → ^AY_Z+1→ ^{A– 4} B* _{Z– 1}→ ^{A– 4} B _{Z– 1},

the particles emitted in the sequence are

(1) γ, β, α

(2) β, γ, α

(3) α, β, γ

(4) β, α, γ

The nucleus ^AX_Z emits a β-particle and its atomic number increases by 1 to transform to the nucleus ^AY_Z+1. The nucleus ^AY_Z+1 emits an α-particle so that its mass number reduces by 4 and atomic number reduces by 2 to become the unstable nucleus ^{A– 4}B* _{Z– 1}. It then emits a γ photon which does not produce any change in mass number and atomic number. The correct option is (4).

2. The number of β-particles emitted by a radioactive substance is equal to twice the number of α -particles emitted by it. The resulting daughter is an

(1) isomer of parent

(2) isotone of parent

(3) isotope of parent

(4) isobar of parent

When an α -particles emitted the mass number reduces by 4 and atomic number reduces by 2. When two β-particles are emitted the atomic number increases by 2. Therefore, when the number of β-particles emitted by a radioactive substance is equal to twice the number of α -particles emitted by it there is no change in the atomic number, but there is a reduction in mass number.

The resulting daughter is therefore an isotope of the parent nucleus [Option (3)].

3. In a Rutherford scattering experiment when a projectile of charge z₁ and mass M₁ approaches a target nucleus of charge z₂ and mass M₂,the distance of closest approach is r₀. The energy of the projectile is

(1) directly proportional to z₁z₂

(2) inversely proportional to z₁

(3) directly proportional to mass M₁

(4) directly proportional to M₁M₂

At the distance of closest approach the entire kinetic energy (E) of the projectile gets converted into electrostatic potential energy of the system. Therefore we have

E = (1/4πε₀) (z₁z₂/r₀)

Therefore, the energy of the projectile is directly proportional to z₁z₂ [Option (1)].