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

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:

• A \(^6Li\) (Lithium-6) loaded plastic scintillator is indeed a highly efficient detector for thermal neutrons.
• When thermal neutrons strike the lithium-6 in the scintillator, an absorption reaction happens where lithium-6 captures the neutron and then disintegrates into an alpha particle (Helium-4) and a triton (Hydrogen-3), releasing a substantial amount of kinetic energy: \(^6Li + ^1n → ^3H + ^4He + 4.78 MeV\)
• Then, this kinetic energy excites the atoms in the scintillating material, causing light to be emitted. This light is then collected and converted into an electrical signal by a photomultiplier tube or other photosensitive device.
• The signal is then amplified and processed. The strength of this signal corresponds to the energy of the incident neutron. Given the high cross-section of the lithium-6 (n, alpha) reaction for thermal neutrons and the relatively high energy of the reaction products, a \(^6Li\)-loaded scintillator presents an efficient option for the detection of thermal neutrons.

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:

The Bethe-Weizsäcker formula:

Where E_B = Binding energy 

The terms on the right-hand side are 

• Volume term: a_vA
• Surface term: - a_sA^2/3  
• Colomb term; - a_c Z(Z-1)/A1/3 
• Asymmetry term and 
• Pairing term: δ(N,Z)

For the most stable atom \( \frac{d(B E)}{d Z}=0\)

EXPLANATION:

\(\begin{aligned} &B . E=15.8 A-18.3 A^{2 / 3}-0.714 \frac{Z(Z-1)}{A^{1 / 3}}-23.2 \frac{(A-2 Z)^2}{A}\\ & ⇒ \frac{d(B E)}{d Z}=0 \\ &\Rightarrow-0.714 \frac{(2 Z-1)}{A^{1 / 3}}+(4 \times 23.2) \frac{(A-2 Z)}{A}=0 \\ & ⇒\frac{92.8}{64}(64-2 Z)=\frac{0.714}{4}(2 Z-1) \\ & ⇒520.89=18.24 Z \quad \Rightarrow Z=28.6 \end{aligned}\)

So, Z/A = 28.6/64 = 0.45

Hence the correct answer is option 3.

CONCEPT:

The nuclear force:

• The nucleus consists of protons and neutrons which are bound by two types of inter-nuclear forces, viz, strong and weak nuclear force. Both of these forces act within the atomic range i.e. within a few fm (1 fm = 10^-15 m). nuclear force is always attractive in nature. 
• The protons have a positive charge but the neutrons are neutral in charge. 
• So, in between the protons a coulombic repulsive force also acts. 

EXPLANATION:

• 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.

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).