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