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 (

*t*

_{2 }–

*t*

_{1}) between the time

*t*

_{2 }when 2/3 of it has decayed and the time

*t*

_{1 }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 = N

_{0}e^{-λt}where N_{0 }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 = N_{0}/2^{n}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
2N

_{0}/3 = N_{0}/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

*t*_{1}= 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
N

_{0}/3 = N_{0}/2^{y}
Therefore, 2

^{y}= 3 so that*y*= (log 3)/(log 2)
Therefore

*t*_{2}= 50(log 3)/(log 2) days
Therefore

*t*_{2 }**–***t*_{1}**= 50 days**
[You may use the decay law N
= N

_{0}e^{-λt}as such to work out the above problem as follows:
At time

*t*_{1}we have
2N

_{0}/3 = N_{0}e^{-}^{λt}^{1}……………(i)
At time

*t*_{2}we have
N

_{0}/3 = N_{0}e^{-}^{λt}^{2}……………..(ii)
From the above equations we have

2
= e

^{λ}^{(}^{t}^{2}^{-}^{t}^{1}^{)}
Therefore

*λ(**t*_{2 }–*t*_{1}) = ℓn 2
Or, (

*t*_{2 }–*t*_{1}) = 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)

*Mc*^{2}–*hν*
(3)

*h*^{2}*ν*^{2}/2*Mc*^{2}
(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*p*^{2}/2M =*h*^{2}*ν*^{2}/2*Mc*^{2}**(3)**A nucleus

_{ n}

*X*

^{m}emits one

*α*–particle and two

*β*–particles. The resulting nucleus is

(1)

_{n}_{–2}*Y*^{m–4}
(2)

_{n}_{–4}*Z*^{m–6}
(3)

_{n}*Z*^{m–6}
(4)

_{ n}*X*^{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

*A*

_{1}and

*A*

_{2}are its activities at times

*t*

_{1}and

*t*

_{2}respectively the number of nuclei which have decayed during the time (

*t*

_{2}–

*t*

_{1}) is

(1)

*A*_{1}*t*_{1}–*A*_{2}*t*_{2}
(2)

*A*_{1}–*A*_{2}
(3) (

*A*_{1}–*A*_{2})/*λ*
(4)

*λ*(*A*_{1}–*A*_{2})
This question appeared in AIPMT Main 2010 question paper.

We have

*N*=*N*_{0}e^{-λt}where*N*_{0 }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 d*N*/d*t =*–*λ**N*_{0}e^{-λt}= –*λ**N*.
The negative sign just inucates that the activity
decreases with time.

Ignoring the negative sign, the activities

*A*_{1}and*A*_{2}at times*t*_{1}and*t*_{2}are given by*A*

_{1}=

*λN*

_{1}and

*A*

_{2}=

*λN*

_{2}where

*N*

_{1}and

*N*

_{2}are the number of nuclei at times

*t*

_{1}and

*t*

_{2}.

Therefore

*N*_{1}=*A*_{1}/*λ*and*N*_{2}=*A*_{2}/*λ*
The number of nuclei which have decayed during the time (

*t*_{2}–*t*_{1}) is*N*

_{1}–

*N*

_{2}=

_{ }(

*A*

_{1}–

*A*

_{2})/

*λ*, as given in option (3).

**(5)**If the nuclear radius of

^{27}Al is 3.6 Fermi, the approximate nuclear radius of

^{64}Cu 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*=*R*_{0}(*A*)^{1/3}where*R*_{0}is a constant an*A*is the mass number.
If

*R*_{1}and*R*_{2}are the nuclear radii of Al and Cu we have*R*

_{1}/

*R*

_{2}= (27/64)

^{1/3}= 3/4

Therefore

*R*_{2}= 4*R*_{1}/3 = (4×3.6)/3 = 4.8 Fermi