## Friday, April 24, 2009

### 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