Kerala Engineering Entrance (KEAM) 2008 Questions on Nuclear Physics

Science without religion is lame, religion without science is blind.

– Albert Einstein

KEAM Engineering 2009 Exam will begin on 25^th of this month. Here are the questions on nuclear physics which appeared in last year’s physics eam:

(1) The nuclear radius of a certain nucleus is 7.2 fm and it has a charge of 1.28×10^–17C.The number of neutrons inside the nucleus is

(a) 136

(b) 142

(c) 140

(d) 132

(e) 126

The nuclear radius R is related to the mass number A as

R = 1.2 A^1/3

On substituting for R in this equation we obtain A^1/3 = 6 so that A = 216.

Since the total charge of the nucleus (carried by all the protons) is given as 1.28×10^–17 coulomb and the charge on a proton is 1.6×10^–19 coulomb, the number of protons (Z) in the nucleus is (1.28×10^–17) /(1.6×10^–19) = 80.

Therefore, the number of neutrons in the nucleus, N = A – Z = 216 – 80 = 136.

(2) The energy released in the fission of 1 kg of ₉₂U²³⁵ is (Energy per fission = 200 MeV)

(a) 5.1×10²⁶ eV

(b) 5.1×10²⁶ J

(c) 8.2×10¹³ J

(e) 5.1×10²³ MeV

1 kg of ₉₂U²³⁵ contains 1/0.235 moles and each mole contains 6.02×10²³ atoms. Therefore1 kg of ₉₂U²³⁵ contains (6.02×10²³)/ 0.235 atoms.

Therefore, the energy E released in the fission of 1 kg of ₉₂U²³⁵ is given by

E = (6.02×10²³) ×200 / 0.235 MeV = 5.1×10²⁶ MeV = 5.1×10³² eV

Since 1 eV = 1.6×10^–19 joule, the above energy is (5.1×10 ³²)×(1.6×10^–19) joule. This works out to approximately 8.2×10¹³ J.

(3) Which one of the following statements is true if half life of a radioactive substance is 1 month?

(a) 7/8^th part of the substance will disintegrate in 3 months

(b) 1/8^th part of the substance will remain undecayed at the end of 4 months

(c) The substance will disintegrate completely in 4 months

(d) 1/16^th part of the substance will remain undecayed at the end of 3 months

(e) The substance will disintegrate completely in 2 months

The number of nuclei N remaining undecayed at the end of n half lives is given by

N = N₀/2ⁿ where N₀ is the initial number.

Therefore, at the end of 3 months (3 half lives in the present case) the number remaining undecayed is N₀/2³ = N₀/8.

This means that 7/8^th part of the substance will disintegrate in 3 months [Option (a)].

You will find a few more multiple choice questions (with solution) on nuclear physics here.

Two EAMCET Questions from Nuclear Physics

The following MCQ which appeared in EAMCET 2005 question paper highlights the law of conservation of momentum as applied to nuclear processes:

A nucleus of mass 218 amu in free state decays to emit an α-particle. Kinetic energy of the α-particle emitted is 6.7 MeV. The recoil energy (in MeV) of the daughter nucleus is

(a) 1

(b) 0.5

(c) 0.25

(d) 0.125

The kinetic energy (K) is given by

K = p²/2m where p is the momentum and m is the mass. The recoil momentum of the daughter nucleus is equal and opposite to the momentum of the α-particle.

We have p²/(2×4 amu) = 6.7 MeV, since the mass of the α-particle is 4 amu.

[You need not convert amu into kilogram and energy into joule and waste your time. You may imagine that p has proper unit to get the energy in MeV]

In the case of the daughter nucleus (of mass 218 – 4 = 214 amu), we have

p²/(2×214 amu) = K, whetre K is the recoil energy of the daughter (in MeV).

From the above two equations we obtain K = 0.125 MeV.

The following MCQ appeared in EAMCET 2000 question paper:

In a nuclear reactor using U²³⁵ as fuel the power output is 4.8 mega watts. The number of fissions per second is (energy released per fission of U²³⁵ = 200 MeV).

(a) 1.5×10¹⁷

(b) 3×10¹⁹

(c) 1.5×10²⁵

(d) 3×10²⁵

This is a very simple question. The energy produced per second is 4.8×10⁶ joule. Since 1 electron volt is 1.6×10^–19 joule, 200 MeV = 200×10⁶×1.6×10^–19 joule.

Therefore, the number of fissions per second = (4.8×10⁶)/(200×10⁶×1.6×10^–19) = 1.5×10^17.

You can find all related posts on this site by clicking on the label ‘nuclear physics’ below this post.