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

Therefore, activity, dN/dt = – λ N₀e^–λt = 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 _ZX^A has mass M kg. If M_p and M_n denote the mass (in kg) of proton and neutron respectively, the binding energy in joule is

(a) [ZM_p + (A – Z)M_n – M]c²

(b) [ZM_p + ZM_n – M]c²

(c) M – ZM_p – (A – Z)M_n

(d) [M– ZM_p – (A – Z)M_n]c²

(e) [AM_n – M]c²

Total mass of the Z protons is ZM_p. 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)M_n.

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 =[ZM_p + (A – Z)M_n – M].

Therefore, binding energy.= ∆Mc² where ‘c’ is the speed of light in free space.

Thus binding energy = [ZM_p + (A – Z)M_n – M]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 = R₀A^1/3 where R₀ is a constant (equal to 1.2×10^–15 m, nearly) and A is the mass number of the nucleus.

If R_l and R₂ are the radii of the given Al and Te nuclei respectively, we have

R_l = R₀ (27)^1/3 = 3R₀ and

R₂ = R₀ (125)^1/3 = 5R₀

Dividing, R_l/R₂ = 3/5

Therefore, R₂ = 5R₁/3 = (5×3.6)/3 fm = 6 fm.

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