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

AIEEE 2010 Questions (MCQ) from Nuclear Physics

“I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend”

– Sir Isaac Newton

Multiple choice questions based on a given paragraph are seen in many entrance exam question papers. Here are two such questions on nuclear physics (which appeared in AIEEE 2010 question paper):

Directions : Questions (i) and (ii) are based on the following paragraph:

A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

(i) The speed of daughter nucleus is

(1) c√[Δm/(M + Δm)]

(2) c [Δm/(M + Δm)]

(3) c√(2Δm/M)

(4) c√(Δm/M)

The energy released in the process is Δmc² in accordance with Einstein’s mass-energy relation. This energy is carried by the daughter nuclei (as kinetic energy).

Therefore we have

½ (M/2)v² + ½ (M/2)v² = Δmc² where v is the speed of each daughter nucleus.

Or, Mv²/2 = Δmc² from which v = c√(2Δm/M), as given in option (3).

[Note that the momentum is to be conserved in the process. Therefore, the daughter nuclei have to travel in opposite directions with the same speed since they have the same mass. If you were asked to give the velocities of the daughters, the answer would be v and – v]

(ii) The binding energy per nucleon for the parent nucleus is E₁ and that for the daughter nuclei is E₂. Then

(1) E₁ = 2 E₂

(2) E₂ = 2 E₁

(3) E₁ > E₂

(4) E₂ > E₁

The binding energy per nucleon for the daughter nuclei is greater than that for the parent nucleus since energy is released in the process[Option (4)].

The following question (MCQ) also was included in the AIEEE 2010 question paper:

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

(1) (A – Z – 4)/(Z – 2)

(2) (A – Z – 8)/(Z – 4)

(3) (A – Z – 4)/(Z – 8)

(4) (A – Z – 12)/(Z – 4)

When an α-particle is emitted, the nucleus loses two protons and two neutrons. When a positron is emitted by the nucleus, a proton in the nucleus gets converted into a neutron. Since the number of neutrons in the original neucleus is A – Z, the number of neutrons in the final nucleus will be (A – Z) – (3×2) + 2 = A – Z – 4.

The protons in the final nucleus will be Z – (3×2) – 2 = Z – 8

Therefore, the ratio of number of neutrons to that of protons in the final nucleus will be(A – Z – 4)/(Z – 8), as given in option (3).

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.

By clicking on the label ‘nuclear physics’ below this post, you can access all posts related to nuclear physics on this site.

You can find useful posts in this section here.

Kerala Engineering Entrance (KEAM) 2008 Questions on Nuclear Physics

Science without religion is lame, religion without science is blind.

– Albert Einstein

KEAM Engineering 2009 Exam will begin on 25^th of this month. Here are the questions on nuclear physics which appeared in last year’s physics eam:

(1) The nuclear radius of a certain nucleus is 7.2 fm and it has a charge of 1.28×10^–17C.The number of neutrons inside the nucleus is

(a) 136

(b) 142

(c) 140

(d) 132

(e) 126

The nuclear radius R is related to the mass number A as

R = 1.2 A^1/3

On substituting for R in this equation we obtain A^1/3 = 6 so that A = 216.

Since the total charge of the nucleus (carried by all the protons) is given as 1.28×10^–17 coulomb and the charge on a proton is 1.6×10^–19 coulomb, the number of protons (Z) in the nucleus is (1.28×10^–17) /(1.6×10^–19) = 80.

Therefore, the number of neutrons in the nucleus, N = A – Z = 216 – 80 = 136.

(2) The energy released in the fission of 1 kg of ₉₂U²³⁵ is (Energy per fission = 200 MeV)

(a) 5.1×10²⁶ eV

(b) 5.1×10²⁶ J

(c) 8.2×10¹³ J

(e) 5.1×10²³ MeV

1 kg of ₉₂U²³⁵ contains 1/0.235 moles and each mole contains 6.02×10²³ atoms. Therefore1 kg of ₉₂U²³⁵ contains (6.02×10²³)/ 0.235 atoms.

Therefore, the energy E released in the fission of 1 kg of ₉₂U²³⁵ is given by

E = (6.02×10²³) ×200 / 0.235 MeV = 5.1×10²⁶ MeV = 5.1×10³² eV

Since 1 eV = 1.6×10^–19 joule, the above energy is (5.1×10 ³²)×(1.6×10^–19) joule. This works out to approximately 8.2×10¹³ J.

(3) Which one of the following statements is true if half life of a radioactive substance is 1 month?

(a) 7/8^th part of the substance will disintegrate in 3 months

(b) 1/8^th part of the substance will remain undecayed at the end of 4 months

(c) The substance will disintegrate completely in 4 months

(d) 1/16^th part of the substance will remain undecayed at the end of 3 months

(e) The substance will disintegrate completely in 2 months

The number of nuclei N remaining undecayed at the end of n half lives is given by

N = N₀/2ⁿ where N₀ is the initial number.

Therefore, at the end of 3 months (3 half lives in the present case) the number remaining undecayed is N₀/2³ = N₀/8.

This means that 7/8^th part of the substance will disintegrate in 3 months [Option (a)].

You will find a few more multiple choice questions (with solution) on nuclear physics here.

IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions on Nuclear Physics

Our greatest weakness lies in giving up. The most certain way to succed is always to try just one more time.

– Thomas Alva Edison

The following three questions [(i), (ii) and (iii)] from nuclear physics were included under Linked Comprehension Type Multiple Choice Questions (single answer type) in the IIT-JEE 2009 question paper. They are simple as you will realize.

Paragraph for Questions (i), (ii) and (iii)

Scientists are working hard to develop unclear fusion reactor. Nuclei of heavy hydrogen, ₁H², known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D–D reaction is ₁H²+₁H² → ₂He³ + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ₁H² nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t₀ before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt₀ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 10¹⁴ s/cm³.

It may be helpful to use the following : Boltzmann constant k = 8.6 × 10^–5 eV/K; e²/4πε₀ = 1.44×10^–9 eVm

Question (i):

In the core of nuclear fusion reactor, the gas becomes plasma because of

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between the deuteron–electron pairs

(D) the high temperature maintained inside the reactor core.

The correct option is (D) since the plasma state is achieved at high temperatures.

Question (ii):

Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10^–15 m is in the range

(A) 1.0×10⁹ K < T < 2×10⁹ K

(B) 2.0×10⁹ K < T < 3×10⁹ K

(C) 3.0×10⁹ K < T < 4×10⁹ K

(D) 4.0×10⁹ K < T < 5×10⁹ K

The total kinetic energy of the two deuterons is 2×1.5 kT = 3 kT. At the required separation r (equal to 4×10^–15 m), the entire kinetic energy gets converted into electrostatic potential energy:

3 kT = e²/4πε₀r

Or, 3×8.6×10^–5×T = 1.44×10^–9/(4×10^–15) since e²/4πε₀ = 1.44×10^–9 eVm

Therefore, T = 1.4×10⁹ K. [Option (A)].

Question (iii):

Results of calculations for four different designs of a fusion reactor using D–D reaction are given below. Which of these is most promising based on Lawson criterion ?

(A) deuteron density = 2.0×10¹² cm^–3, confinement time = 5.0×10^–3 s

(B) deuteron density = 8.0×10¹⁴ cm^–3, confinement time = 9.0×10^–1 s

(C) deuteron density = 4.0×10²³ cm^–3, confinement time = 1.0×10^–11 s

(D) deuteron density = 1.0×10²⁴ cm^–3, confinement time = 4.0×10^–12 s.

Lawson number is nt₀. Out of given options, Lawson number is greater than 5 × 10¹⁴ s/cm³ for option (B) since n = 8.0×10¹⁴ cm^–3 and t₀ = 9.0×10^–1 s.

Therefore the correct option is (B).

You will find some useful multiple choice questions on nuclear physics here