JEE Main Nuclear Physics Previous Year Questions with Solutions

The first real foundation for the concept of the nucleus was provided by Rutherford in 1911. Rutherford while conducting an experiment on the large-angle scattering of alpha particles by matter discovered that all the positive charges of the atom and whole mass of the atom were concentrated in a very tiny central hard core called the nucleus.

Every atomic nucleus contains two types of particles called the neutrons and protons. The protons and neutrons are collectively called the nucleus. The protons are positively charged and the neutrons are neutral. The number of protons in the nucleus of the atom is equal to the electrons in the atom. The number of protons or electrons in an atom is called the atomic number and it is represented as Z. The total number of protons and neutrons in a nucleus is called the mass number, which is represented using the letter A.

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JEE Main Previous Year Solved Questions on Nuclear Physics

Q1: Consider the nuclear fission

Ne²⁰ →2He⁴ + C¹²

Given that the binding energy/ nucleon of Ne²⁰, He⁴ and C¹² are 8.03 MeV, 7.07 MeV and 7.86 MeV respectively. Identify the correct statement.

(a) energy of 9.72 MeV has to be supplied

(b) energy of 12.4 MeV will be supplied

(c) 8.3 MeV energy will be released

(d) energy of 3.6 MeV will be released

Solution

Ne²⁰ →2He⁴ + C¹²

Q – value, E_B = (BE)_initial – (BE)_final

= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12) = 9.72 MeV

Answer: (a) energy of 9.72 MeV has to be supplied

Q2: An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8: 27. The ratio of the radii of the nuclei (assumed to be spherical) is

(a) 3: 2

(b) 2: 3

(c) 4: 9

(d) 8: 27

Solution

As (v₁/v₂) = 8/27; (r₁/r₂) = ?

Using law of conservation of linear momentum, 0 = m₁v₁ – m₂v₂

(As both are moving in opposite directions)

Or m₁/m₂ = v₂/v₁= 27/8

Therefore, (r₁/r₂) = 3/2

Answer: (a) 3: 2

Q3: According to Bohr’s theory, the time-averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the n^th orbit is proportional to (n = principal quantum number)

(a) n^–2

(b) n ^–3

(c) n^–4

(d) n^–5

Solution

Magnetic field at the centre, B_n = μ₀I/2r_n

For a hydrogen atom, the radius of n^th orbit is given by

r_n = (n²/m) (h/2π)²(4πε₀/e²)

r_n ∝ n²

I = e/T = e/(2πr_n/v_n) = ev_n/2πr_n

Also, v_n ∝ n ^–1 ∴ I ∝ n ^–3

Hence, B_n ∝ n^–5

Answer: (d) n^–5

Q4: Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)

(a) 25.8 MeV

(b) 32.4 MeV

(c) 30.2 MeV

(d) 23.6 MeV

Solution

₁H² + ₁H² ➝ ₂He⁴

Energy released = 4(Binding Energy of (₁ ²H)) – 4(Binding Energy of(₂ ⁴He)) = 4 × 7 – 4 × 1.1 = 23.6 MeV

Answer: (d) 23.6 MeV

Q5: As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion

(a) kinetic energy decreases, potential energy increases but the total energy remains the same

(b) kinetic energy and total energy decrease but potential energy increases

(c) its kinetic energy increases but potential energy and total energy decrease

(d) kinetic energy, potential energy and total energy decrease.

Solution

For an electron in n^th excited state of hydrogen atom,

kinetic energy = e²/(8πє₀n²a₀)

potential energy = – e²/(4πє₀n²a₀) and total energy = – e²/(8πє₀n²a₀)

where a₀ is Bohr radius.

As an electron makes a transition from an excited state to the ground state, n decreases. Therefore kinetic energy increases but potential energy and total energy decrease.

Answer: (c) its kinetic energy increases but potential energy and total energy decrease

Q6: Energy required for the electron excitation in Li⁺⁺ from the first to the third Bohr orbit is

(a) 12.1 eV

(b) 36.3 eV

(c) 108.8 eV

(d) 122.4 eV

Solution

Using, E_n = -(13.6Z²/n²) eV

Here, Z = 3 (For Li⁺⁺)

Therefore, E₁ = -(13.6(3)²/1²) = – 122.4 eV

and E₃ =- (13.6(3)²/3²) = – 13.6 eV

ΔE = E₃ – E₁ = – 13.6 + 122.4 = 108.8 eV

Answer: (c) 108.8 eV

Q7: If the binding energy per nucleon in ₃Li⁷and ₂He⁴ nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction

p + ₃Li⁷ → 2 ₂He⁴

energy of proton must be

(a) 39.2 MeV

(b) 28.24 MeV

(c) 17.28 MeV

(d) 1.46 MeV

Solution

Binding energy of ₃Li⁷ = 7 x 5.60 = 39.2 MeV

Binding energy of ₂He⁴ = 4 x 7.06 = 28.24 MeV

Energy of proton = Energy of [2(₂He⁴)- ₃Li⁷]

= 2 × 28.24 – 39.2 = 17.28 MeV

Answer: (c) 17.28 MeV

Q8: A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2: 1. The ratio of their nuclear sizes will be

(a) 2^1/3: 1

(b) 1: 3^1/2

(c) 3^1/2: 1

(d) 1: 2^1/3

Solution

Momentum is conserved during disintegration m₁v₁ = m₂v₂ ——-(1)

For an atom R = R₀A^1/3

(R₁/R₂) = (A₁/A₂)^1/3 = (m₁/m₂)^⅓ = (v₂/v₁)^⅓ from (1)

(R₁/R₂) = (½)^⅓ = 1/2^1/3

Answer: (d) 1: 2^1/3

Q9: A nuclear transformation is denoted by X(n,α)₃Li⁷. Which of the following is the nucleus of element X?

(a) ₅B⁹

(b) ₄Be¹¹

(c) ₆C¹²

(d) ₅B¹⁰

Solution

The nuclear transformation is given by

_zX^A + ₀n¹➝ ₂He⁴ + ₃Li⁷

According to conservation of mass number

A + 1 = 4 + 7 or A = 10

According to conservation of charge number

Z + 0 ➝ 2 + 3 or Z = 5

So the nucleus of the element be ₅B¹⁰.

Answer: (d) ₅B¹⁰

Q10: Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to

(a) 200

(b) 360

(c) 150

(d) 400

Solution

According to law of radioactivity, the count rate at t = 8 seconds is

N₁ = N₀ e^–λt

Taking log on both sides

log N = log N₀  – λt

λt = log N₀ – log N

λt = log (N₀/N )

At t = 0, the number of radioactive particles emitted is 1600

i.e., N₀ = 1600

At t = 8 seconds,  the number of radioactive particles emitted are 100 i.e., N= 100

λ (8) = log (1600/100)

λ (8) = log 16

λ (8) = log 2⁴

λ (8) = 4log 2

λ  = (log 2) /2

We have to find the number of radioactive particles emitting at t = 6 seconds. Therefore,

λt = log (N₀/N )

3 log 2 = log (1600/N)

log 2³= log (1600/N)

log 8 = log (1600/N)

8 = (1600/N)

N = 200

Answer: (a) 200

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