Nuclear and Particle Physics MCQ Quiz in தமிழ் - Objective Question with Answer for Nuclear and Particle Physics - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The atoms of different elements which have the same mass number are called isobars. 

The energy is associated with the electrostatic forces of a system of particles, especially with that of the electrons of a covalent bond.

Explanation:

We are given,

The difference in the Coulomb energy between the mirror nuclei \({ }_{24}^{49} \mathrm{Cr} \text { and }{ }_{25}^{49} \mathrm{Mn}\) is 6.0 MeV.

Again, the radius of the \({ }_{25}^{49} \mathrm{Mn}\) nucleus can be found out by;

\(R=\frac{3 e^2}{5 \cdot \Delta W}\left(Z_1^2-Z_2^2\right)\)

 \(=\frac{3 × 1 × 10^{-15}}{5 × 6}\left(25^2-24^2\right) \) = 4.9 × 10^-15 m.

The correct option is option (2).

Explanation:

We have;

\({ }_{51}^{125} \mathrm{Sb}\) Z = 51 and N = 74

Z = 51.

Using the nuclear shell model, we have:

(s_1/2)² (p_3/2)⁴ (p_1/2)² (d_5/2)⁶ (s_1/2)² (d_3/2)⁴ (f_7/2)⁸ (p_3/2)⁴ (f_5/2)⁶ (p_1/2)² (g_9/2)¹⁰ (g_7/2)¹ 

\(\Rightarrow j=\frac{7}{2}\)and l = 4. Thus spin and parity = \(\left(\frac{7}{2}\right)^{+}\)

\({ }_{38}^{89} \mathrm{Sr}\); Z = 38 and N = 51

N = 51:

(s1/2)2 (p3/2)4 (p1/2)2 (d5/2)6 (s1/2)2 (d3/2)4 (f7/2)8 (p3/2)4 (f5/2)6 (p1/2)2 (g9/2)10 (g7/2)1

\(\Rightarrow j=\frac{7}{2}\) and l = 4. Thus spin and parity = \(\left(\frac{7}{2}\right)^{+}\).

The correct option is option (4).

Explanation:

The reaction can be written in the following way:

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).

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, I_3, 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).

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 + \({1\over 2}\)

<μ_z>_F¹⁹ = μ_N (i + 2.29) = μ_N (2.5 + 2.29) = 4.79 μ_N

\(\rm{ }_{10}^{19}Ne_9\) : 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 + \({1\over 2}\)

<μ_N>_Ne¹⁹ = -(1.91) μ_N 

The correct answer is option (4).

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).

Calculation:

To find the ratio of partial decay widths, we use the square of the relevant Clebsch-Gordan coefficients.

For:

Δ^- → n + π^- : amplitude ∝ <1, –1; 1/2, 1/2 | 3/2, –1/2> = √(1)

Δ⁰ → p + π^- : amplitude ∝ <1, –1; 1/2, –1/2 | 3/2, –3/2> = √(1/3)

So:

Ratio = |Amp_{Δ^- → nπ^-}|² / |Amp_{Δ⁰ → pπ^-}|² = 1 / (1/3) = 3

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).

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:

M_Li v_Li = M_He v_He

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

K_Li = (1/2) M_Li v_Li² = (1/2) M_Li ((M_He v_He) / M_Li)² = (M_He² v_He²) / (2 M_Li)

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

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