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A sample of \({ }_{19} \mathrm{~K}^{40}\) contains invariably \({ }_{18} \mathrm{Ar}^{40} .\) This is because \({ }_{19} \mathrm{~K}^{40}\) has tendency to undergo (a) \(\alpha\) decay (b) positronium decay (c) \(\beta\) decay (d) \(\gamma\) decay

Short Answer

Expert verified
The potassium \(\text{K}^{40}\) undergoes \(\beta\) decay.

Step by step solution

01

Understand the Choices

We need to understand the type of decay processes available: \(\alpha\), positronium, \(\beta\), and \(\gamma\) decay. \(\alpha\) decay involves the emission of an \(\alpha\) particle, which is a helium nucleus. Positronium decay involves the annihilation of an electron and its positron. \(\beta\) decay involves the conversion of a neutron to a proton or vice versa with the emission of an electron or positron. \(\gamma\) decay involves the emission of \(\gamma\) rays, which are high-energy photons.
02

Analyze the Element Change

The transformation in this problem is from \({ }_{19} \mathrm{~K}^{40}\) to \({ }_{18}\mathrm{Ar}^{40}\). This means that the atomic number decreases by 1 (from 19 to 18), while the mass number remains the same (40). This type of change is characteristic of \(\beta^+\) (positron) decay, where a proton is converted into a neutron, leading to a decrease in the atomic number.
03

Conclude the Decay Type

Among the options provided, \(\beta\) decay is the only decay process that can result in a decrease in the atomic number while keeping the mass number constant. Hence, the potassium \({ }_{19} \mathrm{~K}^{40}\) undergoes \(\beta^+\) decay to become \({ }_{18} \mathrm{~Ar}^{40}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Beta Decay
In the world of radioactive decay, beta decay is a fascinating process where the nucleus of an atom undergoes a transformation that changes its fundamental structure. This occurs through two different types of decay: beta-minus (\(\beta^-\)) and beta-plus (\(\beta^+\)) decay.
Let's break these down:
  • Beta-minus decay (\(\beta^-\)): Here, a neutron in the nucleus is transformed into a proton, resulting in the emission of an electron and an antineutrino. This increases the atomic number by 1, while the mass number remains unchanged.
  • Beta-plus decay (\(\beta^+\)): This process changes a proton into a neutron, accompanied by the emission of a positron and a neutrino. Consequently, the atomic number decreases by 1, but again, the mass number stays the same.
Beta decay is integral in altering the identity of an element, allowing atoms to transition from one type to another within the periodic table. What's intriguing is how these changes in the atomic nucleus reflect in the nature of the element, thereby hinting at its stability and its inclination toward decay.
Atomic Number
The atomic number is a basic and very important concept in chemistry and physics. It is denoted by the letter \( Z \) and primarily represents the number of protons present in the nucleus of an atom. This number decisively defines the identity of the element. For instance, all atoms with \( Z = 19 \) are potassium, while those with \( Z = 18 \) are recognized as argon.
In radioactive decay processes such as beta decay, the atomic number often changes:
  • During beta-minus decay: The atomic number increases by 1 as a neutron becomes a proton.
  • During beta-plus decay: The atomic number decreases by 1 as a proton transitions into a neutron.
The atomic number serves not just as an identifier but as a guide to exploring how elements behave in reactions and decay. Understanding changes in the atomic number helps in predicting the outcomes of nuclear reactions and the nature of resulting elements.
Mass Number
The mass number, represented by \( A \), is another essential concept in understanding atoms and their isotopes. It is calculated as the sum of the protons and neutrons in an atom's nucleus, providing insight into the isotopic form of the element.
Unlike the atomic number, the mass number remains constant during beta decay processes:
  • Beta-plus decay: A proton turns into a neutron, but since both are similar in mass, the mass number doesn't change.
  • Beta-minus decay: Here, a neutron is converted into a proton, also resulting in no change to the mass number since they have equivalent masses.
This constancy in the mass number ensures that while the identity of the element may shift due to changes in the atomic number, the overall mass of the atom remains nearly unchanged. The mass number is crucial for understanding isotopic distributions and atomic mass calculations, providing a steady baseline even as nuclear transformations occur.

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Most popular questions from this chapter

The half-life of a radio-isotope is three hours. If the mass of the undecayed isotope at the end of 18 hours is \(3.125 \mathrm{~g}\), what was its mass initially? (a) \(300 \mathrm{~g}\) (b) \(200 \mathrm{~g}\) (c) \(180 \mathrm{~g}\) (d) \(400 \mathrm{~g}\)

\({ }_{92} \mathrm{U}^{2.58}\) emits \(8 \alpha\) particles and \(6 \beta\) particle. The neutron/ proton ratio in the product nucleus is (a) \(60 / 41\) (b) \(62 / 41\) (c) \(61 / 62\) (d) \(61 / 40\)

The number of \(\alpha\) and \(\beta\) particle emitted in the nuclear reaction \({ }^{228} \mathrm{Th}_{90} \longrightarrow{ }^{212} \mathrm{Bi}_{83}\) are (a) \(4 \alpha\) and \(1 \beta\) (b) \(3 \alpha\) and \(7 \beta\) (c) \(8 \alpha\) and \(1 \beta\) (d) \(4 \alpha\) and \(7 \beta\)

The radioisotope, tritium \(\left({ }_{3}^{1} \mathrm{H}\right)\) has a half- life of \(12.3\) years. If the initial amount of tritium is \(32 \mathrm{mg}\), how many milligrams of it would remain after \(49.2\) years? (a) \(4 \mathrm{mg}\) (b) \(8 \mathrm{mg}\) (c) \(1 \mathrm{mg}\) (d) \(2 \mathrm{mg}\)

Select the correct statements: (a) In the reaction \({ }_{11} \mathrm{Na}^{23}+\mathrm{Q} \rightarrow{ }_{12} \mathrm{Mg}^{23}+{ }_{0} \mathrm{n}^{1}\), the bombarding particle \(\mathrm{q}\) is deutron (b) In the reaction \({ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow 56 \mathrm{Ba}^{140}+2\) \({ }_{0} \mathrm{n}^{1}+\mathrm{p}\), produced \(\mathrm{p}\) is \({ }_{36} \mathrm{Kr}^{94}\) (c) In a fission reaction, a loss in mass occurs releasing a huge amount of energy (d) A huge amount of energy is produced during nuclear fission and nuclear fussion reaction

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