Nuclear and Particle Physics MCQ Quiz in తెలుగు - Objective Question with Answer for Nuclear and Particle Physics - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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పొందండి Nuclear and Particle Physics సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Nuclear and Particle Physics MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest Nuclear and Particle Physics MCQ Objective Questions

Top Nuclear and Particle Physics MCQ Objective Questions

Nuclear and Particle Physics Question 1:

A spin-2 nucleus absorbs a spin-1/2 electron and is then observed to decay to a stable nucleus in two stages, recoiling against an emitted invisible particle in the first stage and against an emitted spin-1 photon in the second stage. If the stable nucleus is spinless, then the spin of the invisible particles is?

  1. 1/2, 1
  2. 1/2, 3/2
  3. 3/2, 5/2
  4. 5/2, 7/2

Answer (Detailed Solution Below)

Option 3 : 3/2, 5/2

Nuclear and Particle Physics Question 1 Detailed Solution

Explanation:

The reaction can be written in the following way:
F1 Vinanti Teaching 18.04.23 D8
Here we will check whether the vector sum of the initial and final missing particles spin.

Given in the options gives us possible values of the photon.

If every possibility gives us the photon spin, then that option is correct. 

We have an initial spin of 1/2 and 2 then the possible values of their vector sum are 3/2 and 5/2.

Again, we take the spin of the photon which is 1.

Find the vector sum of those together.

We have now for 3/2 with 1 we get:

1/2, 3/2 and 5/2.

For the 5/2 and 1, we get:

3/2, 5/2, 7/2.

Hence the common possible values for the invisible particle: 3/2, 5/2.

The correct option is (3).

Nuclear and Particle Physics Question 2:

The dominant interactions underlying the following processes,

(a) \( K^- + P \rightarrow \Sigma^- + \pi ^+ \)

(b) \( \mu^- + \mu^+ \rightarrow K^- + K^+ \)

(c) \(\Sigma^+ \rightarrow P + \pi^0 \) are?

  1. EM, EM, strong respectively.
  2. Strong, EM, weak respectively.
  3. Weak, strong, EM respectively.
  4. EM, weak, strong respectively.

Answer (Detailed Solution Below)

Option 2 : Strong, EM, weak respectively.

Nuclear and Particle Physics Question 2 Detailed Solution

Concept:

Strong interactions are responsible for the interactions between nucleons, nucleons and mesons, and a number of other particles.

The mesons act as the quanta of the strong interaction on the nuclear scale.

These interactions reflect the interaction between quarks due to the exchange of gluons on the sub-nucleon level.

Electromagnetic interactions are responsible for the force between electrically charged particles and are mediated by the exchange of photons.

Weak interactions are responsible for many-particle decays such as radioactive decay (the basic process n → p + e- + vg), pion and muon decay, and a number of other decay processes.

Gravitational interactions exist between all particles having mass and are believed to be mediated by the so far undetected gravitons.

Explanation: 

We know that in strong interactions: I, I3, and S all are conserved, where I is the isospin and I3 is the z component of isospin.

In electromagnetic interactions: I is not conserved whereas the S and I3 is conserved.

In weak interactions, the strangeness is not conserved.

In (a) we can check that K-, \(\Sigma^- \)has S = -1 while P and \(\pi^+ \) has S=0.

Hence strangeness is conserved.

Likewise, the I for \(K^+ , P = \frac{1}{2} \) whereas for \(\Sigma^- , \pi^+ = 1 \) .

Hence the reaction (a) obeys the Isospin symmetry.

Thus (a) is obeys strong interactions.

Then for (b) we can see that there are muons (leptons) involved hence this will be either weak or electromagnetic in nature.

For (b)we check strangeness and see that strangeness for \(K^+ , K^- = 1,-1 \) respectively.

whereas for muons it is 0.

Hence strangeness is conserved.

Thus (b) will be electromagnetic.

For (c) we can check that S for Sigma is -1 whereas for protons and Pions it is 0.

Hence the (c) obeys weak interactions.

The correct option is option (2).

Nuclear and Particle Physics Question 3:

The magnetic moments of a proton and a neutron are 2.792 μN and −1.913 μN, where μN is the nucleon magnetic moment. The values of the magnetic moments of the mirror nuclei \(\rm{ }_9^{19}F_{10}\) and \(\rm{ }_{10}^{19}Ne_9\), respectively, in the Shell model, are closest to

  1. 23.652 μN and −18.873 μN
  2. 26.283 μN and −16.983 μN
  3. −2.628 μN and 1.887 μN
  4. 2.628 μN and −1.887 μN

Answer (Detailed Solution Below)

Option 4 : 2.628 μN and −1.887 μN

Nuclear and Particle Physics Question 3 Detailed Solution

Concept:

Magnetic moment, also known as magnetic dipole moment, is the measure of the object's tendency to align with a magnetic field.

Calculation:

\(\rm{ }_9^{19}F_{10}\) : p(9) : 1s\({1\over 2}\)2 1p\({3\over 2}\)2 1p\({1\over 2}\)2 1d\({5\over 2}\)2

j = \({5 \over 2}\)  = 2 + \({1\over 2}\) = l + \({1\over 2}\)

z>F19 = μN (i + 2.29) = μN (2.5 + 2.29) = 4.79 μN

\(\rm{ }_{10}^{19}Ne_9\) : N(9) : 1s\({1\over 2}\)2 1p\({3\over 2}\)4 1p\({1\over 2}\)1 1d\({5\over 2}\)2

j = \({5 \over 2}\)  = l + \({1\over 2}\)

N>Ne19 = -(1.91) μN 

The correct answer is option (4).

Nuclear and Particle Physics Question 4:

In the reaction p + n → p + K+ + X, mediated by strong interaction, the baryon number B, strangeness S and the third component of isospin I3 of the particle X are, respectively

  1. -1, -1 and -1
  2. +1,-1 and -1
  3. +1, -2 and - \(\frac{1}{2}\)
  4. -1,-1 and 0

Answer (Detailed Solution Below)

Option 2 : +1,-1 and -1

Nuclear and Particle Physics Question 4 Detailed Solution

Concept:

Baryons are heavy subatomic particles that are made up of three quarks. Both protons and neutrons, as well as other particles, are baryons. 

Calculation:

 p + n → p + K+ + X

Charge            1    0      1    1   ΔQ = -1

Baryon No.      1    1      1    0   ΔB = 1

Strangeness    0    0      0   -1   ΔS = -1

Hypercharge +\(1\over 2\)  -\(1\over 2\)    +\(1\over 2\)  +\(1\over 2\) ΔI3 = -1 

The correct answer is option (2).

Nuclear and Particle Physics Question 5:

Which of the following elementary particle is a lepton?

  1. Photon 
  2. μ-meson 
  3. π-meson 
  4. Proton 

Answer (Detailed Solution Below)

Option 2 : μ-meson 

Nuclear and Particle Physics Question 5 Detailed Solution

Concept:

Elementary Particle Classification:

Elementary particles are classified into two main categories: fermions and bosons. Fermions include quarks and leptons, while bosons are force carriers.

Leptons are a family of fermions that do not experience the strong nuclear force. They include particles such as electrons, muons, and neutrinos.

 

Identification of Lepton:

The given options are:

1) Photon

2) μ-meson

3) π-meson

4) Proton

Among these options, we need to identify which one is a lepton.

μ-meson (muon) is a lepton. It is similar to the electron but with greater mass.

∴ The correct answer is option 2 (μ-meson).

Nuclear and Particle Physics Question 6:

The nuclear reaction \(n+_{ 5 }^{ 10 }{ B }\rightarrow _{ 3 }^{ 7 }{ Li }+_{ 2 }^{ 4 }{ He }\) is observed to occur even when very slow-moving neutrons \({M}_{n}=1.0087\text{ a.m.u}\) strike a boron atom at rest. For a particular reaction in which \({K}_{n}=0\), the helium \({M}_{He}=4.0026\text{ a.m.u}\) is observed to have a speed of \(9.30× {10}^{6}\text{ m/s}\). \({M}_{Li}=7.0160\text{ a.m.u}\)The kinetic energy of the lithium  __× 10-2 MeV

Answer (Detailed Solution Below) 102

Nuclear and Particle Physics Question 6 Detailed Solution

Calculation:

Since the neutron and boron are both initially at rest, the total momentum before the reaction is zero, and afterward also it is zero. Therefore:

MLi vLi = MHe vHe

We solve this for vLi and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistic formulas, because vHe = 9.30 × 106 m/s is not close to the speed of light (c), and vLi will be even less since vHe = 9.30 × 106 m/s. Thus, we can write:

KLi = (1/2) MLi vLi2 = (1/2) MLi ((MHe vHe) / MLi)2 = (MHe2 vHe2) / (2 MLi)

We put in numbers, changing the mass in u to kg and recalling that 1.60 × 10-13 J = 1 MeV:

KLi = ((4.0026)2 × (1.66 × 10-27) × (9.30 × 106)2) / (2 × (7.0160) × (1.66 × 10-27)) = 1.02 MeV

 

Nuclear and Particle Physics Question 7:

 In the quark model, the state of π+ is given by \(|u\overline d \rangle\). The states for π - and π0 are then given by  

  1. \(|\overline u d \rangle ; \frac{1}{2}|u\overline u+d\overline d\rangle\)
  2. \(|\overline u d \rangle ; \frac{1}{2}|u\overline u-d\overline d\rangle\)
  3. \(|\overline u d \rangle ; |d\overline d\rangle\)
  4. \(|\overline u d \rangle ; |u\overline u\rangle\)

Answer (Detailed Solution Below)

Option 2 : \(|\overline u d \rangle ; \frac{1}{2}|u\overline u-d\overline d\rangle\)

Nuclear and Particle Physics Question 7 Detailed Solution

Concept:

The quark model describes mesons as bound states of a quark and an antiquark. The charged pions ( \(\pi^+\) and\( \pi^- \) ) and the neutral pion ( \(\pi^0\) ) are examples of mesons composed of up ( u ) and down ( d ) quarks, as follows:

Quark states for pions:

  • \(\pi^+\) : Composed of \(|u\bar{d}\rangle\)
  • \(\pi^-\) : Composed of\( |\bar{u}d\rangle\)
  • \(\pi^0\) : A quantum superposition of\( |u\bar{u}\rangle\) and \(|d\bar{d}\rangle ,\) specifically \(\frac{1}{\sqrt{2}}(|u\bar{u}\rangle - |d\bar{d}\rangle) \) , to ensure it is electrically neutral.

Explanation:

  • The \(\pi^+ \) state is straightforward, given as \(|u\bar{d}\rangle \) .
  • The\( \pi^-\) state is its corresponding antiparticle, given by \(|\bar{u}d\rangle\) .
  • The \(\pi^0\) state is a symmetric combination of\( |u\bar{u}\rangle\) and\( |d\bar{d}\rangle\) , with a minus sign for proper symmetry.

Correct Answer: Option 2

\(|u\bar{d}\rangle; \frac{1}{\sqrt{2}}(|u\bar{u}\rangle - |d\bar{d}\rangle)\)

Nuclear and Particle Physics Question 8:

A neutral pi meson has a rest-mass of approximately 140 MeV/c2 and lifetime of τ sec. A  π0 produced in the laboratory is found to decay after 1.25 τ sec. into two photons. Which of the following sets represents a possible set of energies of the two photons as seen in the laboratory? 

  1. 70 MeV and 70 MeV
  2.  35 MeV and 100 MeV
  3. 70 MeV and 100 MeV 
  4. 25 MeV and 150 MeV 

Answer (Detailed Solution Below)

Option 4 : 25 MeV and 150 MeV 

Nuclear and Particle Physics Question 8 Detailed Solution

Concept:

The neutral pion ( \(\pi^0 \) ) decays into two photons. In the laboratory frame, the photons can have different energies depending on the motion of the pion at the time of decay. The total energy of the photons must add up to the total energy of the pion, which includes its rest energy and any additional kinetic energy due to its motion.

Explanation:

  • The rest energy of the \(\pi^0\) meson is approximately 140 MeV ( \(mc^2\) ).
  • If the pion is moving in the laboratory frame, its total energy includes its rest energy and its kinetic energy, leading to an unequal energy distribution among the photons in the decay.

\(E=γ m_0 c^2\)

where m0 =140 MeV/cis the rest mass .

The observed decay time is 

\(t'=γ \tau\).

Thus it is given,

\(t'=1.25 \tau\)

This gives γ =1.25 and

\(\frac{1}{\sqrt{1-v^2/c^2}}=1.25\\ v=0.6c\)

The total energy of neutral pi meson in laboratory frame is 

\(E=γ m_0 c^2\\ E=1.25\times140=175 \ Mev\)

  • In this scenario, the total energy is distributed between two photons as 25 MeV and 150 MeV. This distribution satisfies the energy conservation condition in the laboratory frame.

Correct Answer: Option 4) 25 MeV and 150 MeV

Nuclear and Particle Physics Question 9:

 The four possible configurations of neutrons in the ground state of 4Be9 nucleus, according to the shell model, and the associated nuclear spin are listed below. Choose the correct one: 

  1. (1S1/2)2(1P3/2)3: J=3/2
  2. (1S1/2)2(1P3/2)1(1P1/2)2: J=3/2
  3. (1S1/2)2(1P3/2)4: J=1/2
  4. (1S1/2)2(1P1/2)1(1P3/2)2: J=1/2

Answer (Detailed Solution Below)

Option 1 : (1S1/2)2(1P3/2)3: J=3/2

Nuclear and Particle Physics Question 9 Detailed Solution

Concept:

The nuclear shell model explains the arrangement of nucleons in discrete energy levels or orbitals. The total nuclear spin J depends on the configuration of unpaired nucleons in the valence shell, considering their angular momenta and coupling.

Explanation:

  • 9Be has 4 protons and 5 neutrons. The first 4 neutrons fill the \(1s_{1/2} \) and \(1p_{3/2}\) orbitals completely. The fifth neutron occupies the \(1p_{3/2}\) orbital.
  • The total spin J of the nucleus is determined by the unpaired neutron in the \(1p_{3/2} \) orbital.
  • The \(1p_{3/2}\) orbital has a total angular momentum of j = 3/2 , so the nuclear spin J = 3/2 .

Correct Answer: Option 1

\((1s_{1/2})^2 (1p_{3/2})^3, J = 3/2\)

Nuclear and Particle Physics Question 10:

 The nuclear spins of 14C6 and 25Mg12 nuclei and respectively: 

  1.  zero and half-integer
  2.  half-integer and zero 
  3. an integer and half-integer 
  4. both half-integers

Answer (Detailed Solution Below)

Option 1 :  zero and half-integer

Nuclear and Particle Physics Question 10 Detailed Solution

Concept:

The nuclear spin of a nucleus depends on the total angular momentum of its unpaired nucleons. Nuclei with all nucleons paired (even numbers of protons and neutrons) have zero nuclear spin. For nuclei with unpaired nucleons, the spin corresponds to the angular momentum quantum number of the unpaired nucleon.

Explanation:

  • 14C has 6 protons and 8 neutrons, both even numbers. All nucleons are paired, resulting in zero nuclear spin.
  • 25Mg has 12 protons and 13 neutrons. The last unpaired neutron is in the 1d5/2 orbital, giving it a nuclear spin of 5/2 , which is a half-integer.

Correct Answer: Option 1

Zero and half-integer

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