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 = N0/2x
Therefore, 2x = 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 = N0/2y
Therefore, 2y = 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 nXm
emits one α–particle
and two β–particles.
The resulting nucleus is
(1) n–2Ym–4
(2) n–4Zm–6
(3) nZm–6
(4) nXm–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
