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