AIPMT (Main and Preliminary) Questions on Nuclear Physics

The world is a dangerous place, not because of those who do evil, but because of those who look on and do nothing.

– Albert Einstein

Today we shall discuss a few questions from nuclear physics which appeared in AIPMT question papers. These questions will surely be of use to those who prepare for the National Eligibility Cum Entrance Test (NEET) 2013 for admission to MBBS and BDS Courses. Here are the questions with solution:

(1) The half life of a radioactive nucleus is 50 days. The time interval (t₂ – t₁) between the time t₂ when 2/3 of it has decayed and the time t₁ when 1/3 of it has decayed is

(1) 30 days

(2) 50 days

(3) 60 days

(4) 15 days

This question appeared in AIPMT Main 2012 question paper. You may work it out as follows:

The radioactive decay law is, N = N₀e^-λt where N₀ is the initial number of nuclei, N is the number remaining undecayed after time ‘t’ and λ is the decay constant. This  equation, modified in terms of half life can be written as  N = N₀/2ⁿ where N is number of nuclei remaining undecayed after ‘n’ half life periods.

If 1/3 of the radioactive nucleus decays (and therefore 2/3 of it remains undecayed) in x half life periods, we can write

2N₀/3 = N₀/2^x

Therefore, 2^x = 3/2 so that x = (log 3 – log 2)/log 2 = [{(log 3)/(log 2)} – 1]

The half life of the radioactive nucleus is given as 50 days.

Therefore t₁ = 50[{(log 3)/(log 2)} – 1] days

If 2/3 of the radioactive nucleus decays (and therefore 1/3 of it remains undecayed) in y half life periods, we can write

N₀/3 = N₀/2^y

Therefore, 2^y = 3 so that y = (log 3)/(log 2)

Therefore t₂ = 50(log 3)/(log 2) days

Therefore t₂ – t₁ = 50 days

[You may use the decay law N = N₀e^-λt as such to work out the above problem as follows:

At time t₁ we have

            2N₀/3 = N₀e^-λt1……………(i)

At time t₂ we have

            N₀/3 = N₀e^-λt2……………..(ii)

From the above equations we have

            2 = e ^λ(t2-t1)

Therefore λ(t₂ – t₁) = ℓn 2

Or, (t₂ – t₁) = ℓn 2/λ

But ℓn2/λ is the half life which is 50 days in the present case].     

(2) A radioactive nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be

(1) hν

(2) Mc² – hν

(3) h²ν²/2Mc²

(4) Zero

This question appeared in AIPMT Preliminary 2011 question paper.

The magnitude of the recoil momentum p of the nucleus is the same as that of the photon and is therefore equal to hν/c where c is the speed of light in free space. The kinetic energy of the nucleus is p²/2M = h²ν²/2Mc²

(3) A nucleus  _nX^m emits one α–particle and two β–particles. The resulting nucleus is

(1) _n–2Y^m–4

(2) _n–4Z^m–6

(3) _nZ^m–6

(4) _nX^m–4

This question also appeared in AIPMT Preliminary 2011 question paper.

When an α–particle is emitted the mass number decreases by 4 and the atomic number decreases by 2. When two β–particles are emitted the atomic number increases by 2 but the mass number is unaffected. The resultant nucleus is X itself since the atomic number is unchanged. But it has mass number (m–4). The correct option is (4).

(4) The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at times t₁ and t₂ respectively the number of nuclei which have decayed during the time (t₂ – t₁) is

(1) A₁t₁ – A₂t₂

(2) A₁ – A₂

(3) (A₁ – A₂)/λ

(4) λ(A₁ – A₂)

This question appeared in AIPMT Main 2010 question paper.

We have N = N₀e^-λt where N₀ is the initial number of nuclei, N is the number remaining undecayed after time ‘t’ and λ is the decay constant.

The activity A at time t is dN/dt = – λ N₀e^-λt = – λ N.

The negative sign just inucates that the activity decreases with time.

Ignoring the negative sign, the activities A₁ and A₂ at times t₁ and t₂ are given by

            A₁ = λN₁ and

            A₂ = λN₂ where N₁ and N₂ are the number of nuclei at times t₁ and t₂.

Therefore N₁ = A₁/λ and N₂ = A₂/λ 

The number of nuclei which have decayed during the time (t₂ – t₁) is

            N₁ – N₂ = (A₁– A₂)/λ, as given in option (3).

(5) If the nuclear radius of ²⁷Al is 3.6 Fermi, the approximate nuclear radius of ⁶⁴Cu in Fermi is

(1) 2.4

(2) 1.2

(3) 4.8

(4) 3.6

This question also appeared in AIPMT Preliminary 2012 question paper.

We have nuclear radius R = R₀(A)^1/3 where R₀ is a constant an A is the mass number.

If R₁ and R₂  are the nuclear radii of Al and Cu we have

            R₁/R₂ = (27/64)^1/3 = 3/4

Therefore R₂ = 4R₁/3 =  (4×3.6)/3 = 4.8 Fermi