EmSAT Achieve Exam  >  EmSAT Achieve Tests  >  Physics for EmSAT Achieve  >  Test: Particle Physics - EmSAT Achieve MCQ

Test: Particle Physics - EmSAT Achieve MCQ


Test Description

10 Questions MCQ Test Physics for EmSAT Achieve - Test: Particle Physics

Test: Particle Physics for EmSAT Achieve 2024 is part of Physics for EmSAT Achieve preparation. The Test: Particle Physics questions and answers have been prepared according to the EmSAT Achieve exam syllabus.The Test: Particle Physics MCQs are made for EmSAT Achieve 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Particle Physics below.
Solutions of Test: Particle Physics questions in English are available as part of our Physics for EmSAT Achieve for EmSAT Achieve & Test: Particle Physics solutions in Hindi for Physics for EmSAT Achieve course. Download more important topics, notes, lectures and mock test series for EmSAT Achieve Exam by signing up for free. Attempt Test: Particle Physics | 10 questions in 10 minutes | Mock test for EmSAT Achieve preparation | Free important questions MCQ to study Physics for EmSAT Achieve for EmSAT Achieve Exam | Download free PDF with solutions
Test: Particle Physics - Question 1

A thermal neutron having speed v impinges on a 235⋃ nucleus. The reaction cross-section is proportional to

Detailed Solution for Test: Particle Physics - Question 1

Test: Particle Physics - Question 2

The total angular momentum j of the ground state of the nucleus is

Detailed Solution for Test: Particle Physics - Question 2

The total number of protons here is: z = 8 and the total number of neutrons is: N = 9.
Now, the total angular momentum depends upon the arrangement of the neutrons in the nucleus. The arrangement in the orbits is given below:

The last neutron is in the d-shell with a total angular momentum: 5/2

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Particle Physics - Question 3

When nucleus is bombarded with neutrons, radioactive  and certain particles are produced.

Detailed Solution for Test: Particle Physics - Question 3



Hence, the correct answer is Option-1-protons.

Test: Particle Physics - Question 4

An alpha particle having energy 5 MeV entering into the proportional counter with capacity of 25 pf. The resultant pulse height will be close to:
(Ionisation Potential ~ 15 eV)

Detailed Solution for Test: Particle Physics - Question 4

Pulse height V = Ne/c
Substitute given values, we get,



So, the correct answer is Pulse height V' = 2.133mV

Test: Particle Physics - Question 5

Which of the following decay processes is allowed?

Detailed Solution for Test: Particle Physics - Question 5

​→ For the first process K0 → µ+ + µ-

Thus, this is an allowed decay through the weak interaction. In weak interaction strangeness and
Iso spin is not conserved.
Hence the correct answer is option 1.

Test: Particle Physics - Question 6

Which of the following reactions is not allowed?

Detailed Solution for Test: Particle Physics - Question 6
  • A++ n → ∑0 + + p 
  • Here, "A" represents an arbitrary nucleus with a charge of +2 (A++), "n" represents a neutron, "∑0" represents a neutral sigma particle, and "p" represents a proton.
  • In particle physics and nuclear reactions, there are certain conservation laws that must be obeyed.
  •  Conservation of Charge: The total electric charge before a reaction should equal the total electric charge after the reaction.
  • In this reaction, the initial state (A++ n) has a total electric charge of +2 (A++) for the nucleus and a charge of 0 for the neutron. Therefore, the total initial charge is +2. However, in the final state (∑0 + + p), the neutral sigma particle (∑0) has no charge, and the proton (p) has a charge of +1.
  • This leads to a total electric charge in the final state of +1.
  • Since the total electric charge is not conserved in this reaction (initial state: +2, final state: +1), it is not allowed according to the conservation of charge.
  • So, the correct answer is indeed 2) A++ n → ∑0 + p  because it violates the conservation of charge.
  • The total charge should remain the same before and after a reaction in accordance with the conservation laws of particle physics.
Test: Particle Physics - Question 7

A neutral particle X° is produced in π- + p → X° + n by s-wave scattering. The branching ratios of the decay of X° to 2y, 3π and 2π are 0.38, 0.30 and less than 10-3, respectively. The quantum numbers JCP of X° are

Detailed Solution for Test: Particle Physics - Question 7

In particle physics, the quantum numbers J, C, and P refer to the total angular momentum quantum number, charge parity, and space parity of a particle, respectively. They are extensively used to classify particles and/or to understand their decay channels, and they are fundamentally dictated by the conservation laws and quantum mechanical principles.

  • J (Total angular momentum): A particle decaying into two particles can be in the s-wave state, meaning that the relative angular momentum of the decay products is zero. Thus, the total angular momentum number (J) of the parent particle is equal to the total spin of the decay products. If the particle decays into two photons (2γ), these photons have spin 1 each so J should be an integer (conservation of angular momentum). However, the fact that the particle also decays into 3 pions implies that the angular momentum must be zero (because pions are spin-0 particles and the system is in an s-wave state). Therefore, J = 0.
  • C (Charge parity): All given decay modes allow us to infer that the Charge conjugation must be positive. Therefore, C = +.
  • P (Space parity): Because the decay to two pions (2π) is strongly suppressed (less than 10-3), this means that the parity (P) must be negative. The suppressing of the two-pion decay mode indicates that this mode is probably forbidden by conservation of parity. Therefore, P = -

So, the quantum numbers JCP for the particle X° are 0+-.

Test: Particle Physics - Question 8

The strong nuclear force between a neutron and a proton in a zero orbital angular momentum state is denoted by Fnp(r), where r is the separation between them. Similarly, Fnn (r) and Fpp (r) denote the forces between a pair of neutrons and protons, respectively, in zero orbital momentum state. Which of the following is true on average if the inter-nucleon distance is 0.2 fm < r < 2 fm?

Detailed Solution for Test: Particle Physics - Question 8
  • Inside the nucleus the interaction between neutron-neutron and neutron-proton is
  • always attractive due to nuclear force whereas the force between proton-proton it is repulsive due
  • to coulombic interaction.

Thus, Fnn and Fnp are always attractive and Fpp is repulsive

Hence the correct answer is option 2.

Test: Particle Physics - Question 9

The tensor component of the nuclear force may be inferred from the fact that deuteron nucleus  

Detailed Solution for Test: Particle Physics - Question 9
  • The deuteron, represented by the symbol is a stable isotope of hydrogen with one proton and one neutron, making it a unique and crucial example in nuclear physics.
  • The existence of the tensor force can be inferred from the properties of the deuteron. The deuteron is the only stable particle composed of two nucleons, and yet, it is not particularly tightly bound, reflecting the balance between the attractive strong force and the repulsive coulomb and centrifugal forces.
  • In quantum mechanics, particles can have different types of spin states. In the case of a system of two nucleons, such as the deuteron, there can be a triplet state (where the spins of the two nucleons are aligned, giving a total spin of 1) and a singlet state (where they are anti-aligned, giving a total spin of 0).
  • The fact that the deuteron exists in a triplet state (spin = 1), but a neutron-neutron or proton-proton spin-singlet bound state does not exist, strongly suggests the existence of a tensor component in the nuclear force.
  • This is because the tensor force can couple to spin-1 states but not to spin-0 states. Without the tensor force, the deuteron would not be bound.
  • A non-zero quadrupole moment in the ground state of the deuteron is only possible if there's a combination of both S and D states (where D state signifies a quantum mechanical state with an orbital angular momentum quantum number l = 2).
  • The mixing of S and D states is a clear signature of the tensor component of the nuclear force. So the deuteron's non-zero electric quadrupole moment in its ground state works as indirect evidence for the tensor component of the nuclear force.
Test: Particle Physics - Question 10

The Q - value of the α - decay of 232Th to the ground state of 228Ra is 4082 keV. The maximum possible kinetic energy of the α - particle is closest to

Detailed Solution for Test: Particle Physics - Question 10


The correct answer is option (4).

208 videos|230 docs|191 tests
Information about Test: Particle Physics Page
In this test you can find the Exam questions for Test: Particle Physics solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Particle Physics, EduRev gives you an ample number of Online tests for practice

Top Courses for EmSAT Achieve

208 videos|230 docs|191 tests
Download as PDF

Top Courses for EmSAT Achieve