Saturday, November 21, 2009

Questions (MCQ) on Nuclear Physics

The following three questions from nuclear physics are simple but are useful in your preparation for entrance tests:
(1) Two radioactive samples S1 and S2 have half lives 3 hours and 7 hours respectively. If they have the same activity at a certain instant t, what is the ratio of the number of atoms of S1 to the number of atoms of S2 at the instant t?
(a) 9 : 49
(b) 49 : 9
(c) 3 : 7
(d) 7 : 3
(e) 1 : 1
If The number of atoms present at the instant t is N, we have
N = N0eλt where N0 is the initial number, e is the base of natural logarithms and λ is the decay constant.
Therefore, activity, dN/dt = λ N0eλt = λN
If N1 and N2 are the number of atoms of S1 and S2 respectively when the activities are the same, we have
λ1N1 = λ2N2 from which N1/N2 = λ2/λ1
But the decay constant λ is related to the half life T as T = 0.693/λ.
Therefore, N1/N2 = λ2/λ1 = T1/T2 = 3/7 [Option (c)].
(2) A nucleus ZXA has mass M kg. If Mp and Mn denote the mass (in kg) of proton and neutron respectively, the binding energy in joule is
(a) [ZMp + (A – Z)MnM]c2
(b) [ZMp + ZMnM]c2
(c) M – ZMp – (A – Z)Mn
(d) [M– ZMp – (A – Z)Mn]c2
(e) [AMn M]c2
Total mass of the Z protons is ZMp. Since the total number of nucleons is A, the total number of neutrons is (A – Z) and the total mass of the neutrons is (A – Z)Mn.
The mass defect M is the difference between the total mass of the nucleons (protons and neutrons together) and the mass of the nucleus: M =[ZMp + (A – Z)MnM].
Therefore, binding energy.= Mc2 where ‘c’ is the speed of light in free space.
Thus binding energy = [ZMp + (A – Z)MnM]c2
(3) If the aluminium nucleus 13Al27 has nuclear radius of about 3.6 fm, then the tellurium nucleus 52Te125 will have radius approximately equal to
(a) 3.6 fm
(b) 16.7 fm
(c) 8.9 fm
(d) 6.0 fm.
(e) 4.6
The nuclear radius R is given by
R = R0A1/3 where R0 is a constant (equal to 1.2×10–15 m, nearly) and A is the mass number of the nucleus.
If Rl and R2 are the radii of the given Al and Te nuclei respectively, we have
Rl = R0 (27)1/3 = 3R0 and
R2 = R0 (125)1/3 = 5R0
Dividing, Rl/R2 = 3/5
Therefore, R2 = 5R1/3 = (5×3.6)/3 fm = 6 fm.
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