Nuclear Chemistry MCQ Quiz in मल्याळम - Objective Question with Answer for Nuclear Chemistry - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is 199mL

Concept:-

• Radioactive Tracers: In biology and chemistry, a radioactive tracer, or radioactive label, is a substance that contains a radioisotope, so its movements can be traced using a radiation detection device. In this case, we are injecting a radioactive isotope into the animal's bloodstream to measure the total volume.
• Equilibrium Distribution: Once injected into the bloodstream, the tracer would distribute evenly throughout the animal's blood. The assumption here is that sufficient time has passed for circulatory equilibrium (uniform distribution of the tracer) to occur.
• Dilution Principle: The principle of dilution is used to solve this problem. The dilution factor is defined as the total final volume of the diluted solution per the volume of the extracted substance. Despite diluting a substance, the total amount remains the same as long as the solution is well mixed.
• Radioactivity and Decay per Minute (dpm): This is a measure of radioactivity. One decay per minute is equal to one atom undergoing decay each minute.

Explanation:-

No change in the activity of the sample during the establishment of circulatory equilibrium.
Let volume of blood is V mL,

so total vol.
= (V + 1) mL after injection of sample.
2 mL sample has activity of 10 dpm

so (V+1) mL sample has activity of 10/2 x (V+1)
Since rate is constant so
10/2 × (V + 1) = 1000;

V = 199 mL

Conclusion:-

So the volume is 199ml

Concept:

Nuclear decay, also known as radioactive decay, is a natural and spontaneous process by which the nucleus of an unstable atom transforms into a more stable configuration by emitting particles or energy. This process occurs in certain types of atomic nuclei, particularly those with too many or too few protons and neutrons to be in a stable state. Beta decay is one of the primary type of nuclear decay. Beta decay involves the transformation of a neutron into a proton or a proton into a neutron within the nucleus. This process leads to the emission of either a beta-minus (β-) particle (an electron) or a beta-plus (β+) particle (a positron), depending on the type of beta decay. Beta decay results in a change in atomic number but no change in mass number. There are two primary types of beta decay:

Beta-minus decay:

In beta-minus decay, a neutron in the nucleus is transformed into a proton, an electron (often referred to as a beta-minus particle), and an antineutrino (ν̅e). The process can be represented as follows:

n (neutron) → p (proton) + e^- (beta-minus particle) + ν̅e (antineutrino)

The emitted beta-minus particle (e^-) carries away some of the energy released in the decay process.

Beta-positive decay:

In beta-plus decay, a proton in the nucleus is transformed into a neutron, a positron (often referred to as a beta-plus particle), and a neutrino (νe). The process can be represented as follows:

p (proton) → n (neutron) + e⁺ (positron) + νe (neutrino)

The emitted positron (e⁺) is the antimatter counterpart of the electron (e^-) and carries away some of the energy released in the decay process.

Explanation:

• When ¹⁹Ne emits a positron, it undergoes beta plus decay. In this process, a proton in the nucleus is transformed into a neutron, and a positron (a positively charged electron) is emitted, resulting in a change of the atomic number. So, ¹⁹Ne becomes ¹⁹F. The reaction can be represented as follows:

• When ¹⁴C emits a beta particle (β^- particle), it undergoes beta minus decay, which is also known as electron emission. In this process, a neutron in the nucleus is transformed into a proton, and a beta minus particle or electron (denoted as β^-) is emitted ,which increases its atomic number. So, ¹⁴C becomes ¹⁴N. The reaction can be represented as follows:

Conclusion:

Therefore, the resulting species are ¹⁹F and ¹⁴N.

Concept:-

• For a given element, the decay or disintegration rate is known as specific activity and it is proportional to the number of atoms.
• The activity is measured in terms of the number of disintegrations per unit of time.
• The relation between decay or disintegration rate with the number of atoms is given by,

\( - {{dN} \over {dt}} = \lambda N\)

• Where, \({{dN} \over {dt}}\) is the decay or disintegration rate, is the disintegration constant, and N is the number of atoms present at time t.
• For a first-order reaction, the half-life (t1/2) is related to the disintegration constant () as,

​\({t_{{1 \over 2}}} = {{0.693} \over \lambda }\)

Explanation:-

• The atomic weight of ³¹P is 30.97 gm.
• 30.97 gm of phosphorus contains 6.022 × 10²² atoms.
• Thus, 1000 gm of phosphorus contains

\(=\frac{6.022\times 10^{23}\times 1000}{30.97}\) atoms

= 1.94 × 10²⁵ atoms.

• Now, the specimen contains 1 part per million ³²P is,

= 1.94 × 1025 × 10^-6 atoms

= 1.94 × 10¹⁹ atoms

• The value of the disintegration constant (\(\lambda \)) is,

​\(\lambda = \frac{0.693}{t_{\frac{1}{2}}}\) s-1

\(\lambda = \frac{0.693}{14.3\times 24\times 60\times 60}\) s^-1

• Thus, the specific activity of a specimen of phosphorus-containing 1 part per million 32P is,

\(A = \lambda\times n = \frac{0.693}{14.3\times 24\times 60\times 60} \times 1.94 \times 10^{19}\) s-1

= 1.088 × 10¹³ s^-1

\(=\frac{1.088\times 10^{23}}{3.7\times 10^{10}}\) Ci.kg-1

= 294 Ci.kg-1

​Conclusion:-

• Hence, the specific activity of a specimen of phosphorus-containing 1 part per million 32P is = 294 Ci.kg-1.

Concept:

→ K-electron capture is a type of radioactive decay process in which an unstable atomic nucleus captures one of its own inner-shell electrons (usually a K-electron) and combines with a proton in the nucleus to form a neutron. This process results in a decrease in atomic number by one and no change in mass number.

Explanation:

→ The K-electron capture occurs when the binding energy of an electron in the K-shell is greater than the energy difference between the initial and final states of the nucleus.

→ When this condition is met, the electron can be captured by the nucleus and combines with a proton to form a neutron. The process is represented by the following equation:

p + e^- → n + ν

where p represents a proton, e- represents a K-shell electron, n represents a neutron, and ν represents an electron neutrino that is emitted during the process.

→ K-electron capture is a type of radioactive decay process in which an unstable nucleus captures one of its own inner-shell electrons (usually a K-electron) and combines with a proton in the nucleus to form a neutron. This process results in a decrease in atomic number by one and no change in mass number.

→ During K-electron capture, the nucleus emits a neutrino (ν) in order to conserve lepton number. Lepton number is a fundamental quantity in particle physics that is conserved in all interactions. The electron neutrino (ν_e) is the type of neutrino that is typically emitted during K-electron capture.

Conclusion:

The correct answer is A and B.

Concept:

Nuclear reaction:-

• A nuclear reaction is a process in which two nuclei, or more nuclei and other external subatomic particles collide with each other and form one or more new nuclei.
• α (alpha) particle is a positively charged particle with 2 protons and 2 neutrons. It is represented as 2He4 or 2α4.
• β (beta) particle is negatively charged. It is represented as -1e0.

Explanation:

• The atomic number of Bi is 83.
• The emission of an α particle results in the decrease of mass number by 4 units and atomic number by 2 unit. The mass number and the atomic number of resulting nuclei will be,

\(\left( {212 - 4} \right) = 208\), and

\(\left( {83 - 2} \right) = 81\)

• When a 212Bi decays with the emission of an α particle the nuclear reaction will be:

​\(^{{\rm{212}}}{\rm{B}}{{\rm{i}}_{{\rm{83}}}}{ \to ^{{\rm{208}}}}{{\rm{X}}_{{\rm{81}}}}{{\rm{ + }}^{\rm{4}}}{{\rm{\alpha }}_{\rm{2}}}\)

• The release of a β particle will increase the atomic number by one unit, whereas the mass number will be unchanged. So, the mass number and the atomic number of resulting nuclei will be,

208, and

\(\left( {81 + 1} \right) = 82\)

• When a 212Bi decays with the emission of an α particle the nuclear reaction will be:

\(^{{\rm{208}}}{\rm{X}_{{\rm{81}}}}{ \to ^{{\rm{208}}}}{{\rm{Pb}}_{{\rm{82}}}}{{\rm{ + }}^{\rm{0}}}{{\rm{\beta }}_{\rm{-1}}}\)

Conclusion:

Hence, the element will be Pb.

The correct answer is Critical mass

Concept:-

• Nuclear Chain Reaction: Describes the process where one nuclear reaction triggers additional reactions, leading to a self-sustaining cascade.
• Critical Mass: The minimum amount of fissile material required for a sustained chain reaction.
• Neutron Moderation: Involves using materials to slow down fast neutrons, increasing their probability of causing fission reactions.
• Flux: Represents the rate of flow of neutrons, a critical parameter in achieving and maintaining criticality.

Explanation:-

Critical mass is the minimum amount of fissile material needed to sustain a self-sustaining nuclear chain reaction. It marks the point at which the rate of neutron production equals the rate of neutron loss.

Chain Reaction Fundamentals:

• A nuclear chain reaction involves the splitting (fission) of fissile nuclei, releasing energy and neutrons.
• Neutrons produced during fission can induce further fission reactions in neighboring fissile nuclei, creating a self-sustaining process.

Influence of Factors on Critical Mass:

• Factors affecting critical mass include fissile material purity, shape, density, and the presence of reflectors or moderators.
• Certain shapes and configurations can either increase or decrease the critical mass required.

Role of Flux and Neutron Moderation:

• Flux, representing the rate of neutron flow, is crucial for achieving and maintaining criticality.
• Neutron moderators, such as water or graphite, can slow down fast neutrons, making them more effective in causing fission reactions and reducing the critical mass.

Conclusion:-

So, The minimum amount of fissionable material that can produce a self sustaining chain reaction is called Critical mass

Concept:

Carbon-14 (¹⁴C) is a radioactive isotope of carbon, commonly used in radiocarbon dating to determine the age of ancient artifacts and geological samples. The decay of ¹⁴C is a first-order nuclear process where it emits particles to transform into a stable isotope.

• Radioactive Decay: Involves the emission of particles to achieve a more stable nuclear configuration.
• Beta Decay: This is a type of radioactive decay where a neutron is converted into a proton, and the nucleus emits a beta particle (an electron or positron).

Explanation:

In the case of carbon-14 dating, ¹⁴C undergoes beta decay, converting a neutron into a proton and emitting a beta particle (an electron) in the process. The reaction can be represented as follows:

• Nuclear reaction for ¹⁴C decay:

  • \( ^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + β^- + \bar{\nu}_e \)
    Here, β^- represents the beta particle (electron) emitted, and \(\bar{\nu}_e \) represents the antineutrino.

Conclusion:

In carbon dating, ¹⁴C emits a β particle. Therefore, the correct answer is  β particle.

Concept:

In nuclear chemistry, magic numbers are the numbers of nucleons (either protons or neutrons) that are arranged into complete shells within the atomic nucleus. These numbers correspond to especially stable configurations. The most commonly recognized magic numbers are:

• Magic Numbers: 2, 8, 20, 28, 50, 82, and 126

Explanation:

• 28 is a magic number (nuclear shell model).
• 20 is a magic number (nuclear shell model).
• 120 is not a magic number.
• 50 is a magic number (nuclear shell model).

Conclusion:

Among the given options, the number that is not a magic number is 120.

Concept:

Induced radioactivity refers to nuclear reactions in which a stable nucleus becomes radioactive upon bombardment with a subatomic particle or another nucleus. This process often involves neutron capture or particle-induced reactions, resulting in a new isotope or element that may be unstable and undergo radioactive decay.

Explanation: 

• Reaction (a): ¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H
  This reaction involves the bombardment of nitrogen with an alpha particle (He), resulting in the formation of oxygen and a proton. This is an example of induced radioactivity as it produces a new element, oxygen, through a nuclear reaction.

• Reaction (b): ⁹₄Be + ¹₁H → ⁶₃Li + ⁴₂He
  In this reaction, beryllium is bombarded by a proton, resulting in lithium and an alpha particle. This reaction is also considered an example of induced radioactivity due to the transformation of one element (Be) into another (Li).

• Reaction (c): ²⁴₁₂Mg + ⁴₂He → ²⁷₁₄Si + ¹₀n
  Here, magnesium is bombarded with an alpha particle, producing silicon and a neutron. This is another example of induced radioactivity, as the process results in the formation of silicon and neutron emission.

• Reaction (d): ¹⁰₅B + ⁴₂He → ¹³₇N + ¹₀n
  This reaction shows boron being bombarded by an alpha particle, resulting in nitrogen and a neutron. It also exemplifies induced radioactivity as a stable boron nucleus transforms into a different nucleus (N).

Conclusion:

The correct answer is all reactions (a, b, c, and d) as each represents an example of induced radioactivity.

The correct answer is 11

Explanation:-

\({ }_{92}^{238} U \rightarrow{ }_{82}^{210} \mathrm{~Pb}\)

for α - decay the atomic mass reduces by 4 units and atomic number reduces by 2 unit

for β - decay the atomic number increases by 1 unit

let x be the number of α - decay

and, y be the number of β - decay

then,

from atomic mass → 238 - 4x = 210x = 7$

from atomic number → 92 - 2x + y = 82

put x = 7y = 4

x + y = 7 + 4 = 11

So, total α & β particle emitted are 11 .