Kerala Medical Entrance (KEAM) 2008 Questions on Nuclear Physics

Try not to be a person of success, but rather a person of virtue.

– Albert Einstein

Here are the two questions from nuclear physics which were included in KEAM (Medical) 2008 question paper:

(1) If the mass defect of ₈O¹⁶ nucleus is 0.128 amu, then the binding energy per nucleon of oxygen is

(a) 8.2 MeV

(b) 7.45 MeV

(c) 7.3 MeV

(d) 7.1 MeV

(e) 8.15 MeV

One atomic mass unit (amu) is equivalent to 931 MeV. Therefore, the total binding energy of the ₈O¹⁶ nucleus is 0.128×931 MeV.

Since there are 16 nucleons in the ₈O¹⁶ nucleus, the binding energy per nucleon of oxygen is (0.128×931)/16 = 7.45 MeV, very nearly.

(2) Two radioactive samples have decay constants 15x and 3x. If they have the same number of nuclei initially, the ratio of number of nuclei after a time 1/6x is

(a) 1/e

(b) e/2

(c) 1/e⁴

(d) 2e/3

(e) 1/e²

The number N of nuclei at time t is given by

N = N₀e^–λt where N₀ is the initial number, e is the base of natural logarithms, and λ is the decay constant.

The required ratio is (N₀e^–15x/6x)/ (N₀e^–3x/6x) = e^–2.5/ e^–0.5 = e^–2 = 1/e²

Three questions from nuclear physics were included in the physics question paper of Kerala Engineering Entrance (KEAM) 2008 examination. You will find those questions with solution here

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions on Nuclear Physics

Here are two multiple choice questions which appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008:

(1) Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively. If initially they have the same number of nuclei then the ratio of the number of nuclei of X₁ to that of X₂ will be 1/e after a time

(1) λ/2

(2) 1/(4λ)

(3) e/λ

(4) λ

If the initial number of nuclei is N₀ we have

N₁ = N₀e^{–5λ t} and

N₂ = N₀e^{–λ t} where N₁ and N₂ are the number of nuclei of X₁ and X₁₂ at time t.

Therefore N₁/N₂ = e^–4λt

This will be equal to 1/e when t =1/(4λ).

(2) Two nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be

(1) 3:1

(2) (3)^1/3:1

(3) 1:1

(4) 1:3

The mass of a nucleus is directly proportional to the number (A) of the nucleons. The volume of the nucleus is (4/3)πR³ where R is the nuclear radius. But, R = R₀A^⅓ where R₀ is a constant (equal to 1.1 ×10^-15m). So, the volume of the nucleus also is directly proportional to the nucleon number A. Since the density is the ratio of mass to volume, it follows that the density of nuclear matter is independent of the nucleon number A so that the correct option is (c).

You will find some useful multiple choice questions (with solution) on nuclear physics at physicsplus

Multiple Choice Questions on Nuclear Fission

Multiple choice questions on nuclear fission at the level expected from you will be simple. Here are three such questions:

(1)The energy released per fission of a U²³⁵ nucleus is around

(a) 0.02 eV

(b) 2 eV

(c) 2 MeV

(d) 20 MeV

(e) 200 MeV

The mass of the products of fission will be less than the mass of the U²³⁵ nucleus. The mass difference gets converted into energy. Even though different fragments can be produced, the energy released per fission of a U²³⁵ nucleus is around 200 MeV.

(2) From the following, pick out the most suitable energy of neutrons which will produce nuclear fission in a reactor

(a) 0.04 eV

(b) 40 eV

(c) 400 eV

(d) 2 MeV

(e) 20 MeV

Nuclear fission is induced most effectively by neutrons of thermal energy and hence the correct option is 0.04 eV

(3) The function of the moderator in a nuclear reactor is

(a) to absorb fast neutrons

(b) to adjust the power output to moderate levels

(c) to slow down fast neutrons

(d) to absorb slow neutrons

(e) to cool the reactor core

This is a simple question which repeatedly appears in entrance test papers with minor changes in the wrong options. The moderator is used to slow down fast neutrons to thermal energies [Option (c)].

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

You can find some useful multiple choice questions (with solution) from nuclear physics at physicsplus

Two Kerala Engineering Entrance 2007 Questions from Nuclear Physics

The following MCQ (on radioactivity) which appeared in Kerala Engineering Entrance 2007 question paper is simple, but it differs slightly from the conventional type:

Radium has half life of 5 years. The probability of decay of a radium nucleus in 10 years is

(a) 50% (b) 75% (c) 100% (d) 60% (e) 25%

Since the half life is 5 years, the amount getting decayed in 10 years will be 75%.

[This can be found very easily: After 5 years half the initial amount will be decayed; after another 5 years, half of the remaining will be decayed. If the half life and the time period given are not so simply related, you will have to calculate the number of nuclei undecayed (N) using the equation, N = N₀/2ⁿ where N₀ is the initial number and ‘n’ is the number of half lives in the given time. The percentage decayed in the given time is then calculated].

The probability for decay of any given nucleus in 10 years is therefore 75%.

The following question involving the relative abundance of isotopes is a popular one and it has found place in Kerala Engineering Entrance 2007 question paper:

The natural boron of atomic weight 10.81 is found to have two isotopes B¹⁰ and B¹¹. The ratio of abundance of isotopes in natural boron should be

(a) 11:10(b) 81:19 (c) 10:11 (d) 15:16 (e) 19:81

This is a simple arithmetical problem. If there are n₁ boron atoms of atomic weight 10 and n₂ boron atom of atomic weight 11 in a sample of natural boron, the mean atomic weight (which is given as 10.81) is related to n₁ and n₂ as

(10n₁ + 11n₂)/ (n₁ + n₂) = 10.81.

Rearranging, 0.81n₁ = 0.19n₂, from which n₁/n₂ = 19/81.

Radioactive Decay Law

The radioactive decay law as you might be remembering well is expressed mathematically as
N = N0e^-λt with usual notations.
In most entrance examinations such as Medical and Engineering entrance examination, you wont be allowed to use calculators or logarithm tables. The above equation, modified in terms of half life will be very useful in this context. If N is the number of nuclei remaining undecayed after ‘n’ half life periods, it is related to the initial number N0 as,
N = N0/2ⁿ.
Now let us discus the following M.C.Q.:
Out of 1.414×10²⁴ nuclei, only 10²⁴ nuclei remain undecayed after 15 minutes in a radioactive sample. The half life period of the sample in minutes is
(a) 64 (b) 55 (c) 40 (d)30 (e) 24
We have, 10²⁴ = (1.414×10²⁴)/2ⁿ from which 2ⁿ = 1.414 so that n= ½. This means that 15 minutes is half of the half life period. The half life of the sample therefore is 30 minutes.
The above question can be asked in a modified manner, involving the activity of the sample as follows:
The activity of a radioactive sample drops from 1.414×10⁸ disintegrations per second to 10⁸ disintegrations per second in 15 minutes. The half life period of the sample in minutes is
(a) 64 (b) 55 (c) 40 (d)30 (e) 24

Since the activity of a sample is directly proportional to the number of nuclei present at the instant, we can express the activity ‘A’ after ‘n’ half lives in terms of the initial activity ‘A0’ as,
A = A0/2ⁿ
Substituting the values of A and A0, we have 10⁸ = (1.414×10⁸)/2ⁿ from which n=½. So 15 minutes is half of the half life period of the sample and the answer to the question is 30 minutes [option (d)].