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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

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The decay is (b) positronium decay.

Step by step solution

01

Understanding the Scenarios

We need to determine which type of decay causes \( _{19} \mathrm{K}^{40} \) to transform into \( _{18} \mathrm{Ar}^{40} \). Each decay option affects atomic and mass numbers differently, so recognizing their specific effects is crucial.
02

Identifying Alpha Decay

In alpha decay, an alpha particle (consisting of 2 protons and 2 neutrons) is emitted from the nucleus. This decreases the atomic number by 2 and mass number by 4.
03

Considering Positron Emission

During positron emission, a proton in the nucleus is converted into a neutron, emitting a positron. This decreases the atomic number by 1, with no change to the mass number.
04

Evaluating Beta Decay

In beta decay, a neutron is converted to a proton, and a beta particle (electron) is emitted. This increases the atomic number by 1 without affecting the mass number.
05

Analyzing Gamma Decay

Gamma decay involves emission of gamma radiation, resulting in no change to the atomic number or mass number. It only releases excess energy from the nucleus.
06

Determining the Correct Decay

\( _{19} \mathrm{~K}^{40} \) transforms to \( _{18} \mathrm{Ar}^{40} \) by a decrease in the atomic number (by 1) with constant mass number. This is consistent with positron emission or electron capture, where a proton changes to a neutron.

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Key Concepts

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

Alpha Decay
Alpha decay is a type of radioactive decay where an unstable nucleus releases an alpha particle. An alpha particle consists of two protons and two neutrons, similar to a helium nucleus. When a nucleus undergoes alpha decay, it loses these four nucleons.
This results in a decrease of the atomic number by 2 and the mass number by 4. To better understand the impact of alpha decay, imagine a heavier nucleus like uranium ( ^{238} ext{U} ) undergoing this process. It transforms into thorium ( ^{234} ext{Th} ) because:
  • The atomic number reduces from 92 to 90 (losing 2 protons).
  • The mass number decreases from 238 to 234 (losing 4 nucleons).
Alpha particles have a relatively low penetrative power. They can typically be stopped by a sheet of paper or even the outer layer of human skin.
However, if inhaled or ingested, they can cause significant damage due to their high ionizing capability.
Positron Emission
Positron emission is a fascinating process where a proton in the nucleus of an atom is transformed into a neutron. During this transformation, a positron is emitted. A positron is the antimatter counterpart of an electron: it has the same mass but a positive charge.
In positron emission, the atomic number of the nucleus decreases by 1, but the mass number remains unchanged. As a result, one element changes into another with one fewer proton. For instance, consider potassium-40 ( ^{40} ext{K} ), which undergoes positron emission to become argon-40 ( ^{40} ext{Ar} ):
  • The atomic number changes from 19 (potassium) to 18 (argon), as one proton converts to a neutron.
  • The mass number remains at 40, with the total nucleons unchanged.
This transformation explains why a sample of ^{40} ext{K} is often found with ^{40} ext{Ar} due to positron emission.
Although not as ionizing as alpha particles, emitted positrons quickly meet electrons and annihilate each other, releasing energy in the form of gamma radiation.
Gamma Decay
Gamma decay is a process where a nucleus in an excited state releases energy in the form of gamma radiation. Unlike alpha or beta decay, gamma decay does not change the number of protons or neutrons in the nucleus. This means both the atomic number and mass number remain constant. Gamma rays are highly penetrating electromagnetic waves. They are similar to X-rays but generally have higher energy. These rays can pass through most materials and require thick lead or concrete barriers to be effectively blocked. Despite not changing the element or isotope, gamma decay is crucial in reducing the energy state of a nucleus post other types of decay. For example, when a nucleus undergoes alpha or beta decay, it might be left in an excited state due to the rearrangement of nucleons.
Through gamma decay, the nucleus can release this excess energy and settle into a more stable state. It's important to note that while gamma decay doesn't change elemental identities, the release of high-energy photons is significant, especially in fields like medical imaging and radiation therapy.

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Most popular questions from this chapter

Two radioactive elements A and B have decay constant \(\lambda\) and \(10 \lambda\) respectively. If the decay begins with the same number of atoms of the \(\mathrm{n}\), the ratio of atoms of \(\mathrm{A}\) to those of B after time \(1 / 9 \lambda\) will be (a) \(\mathrm{e}^{-3}\) (b) \(\mathrm{e}^{2}\) (c) \(\mathrm{e}\) (d) \(\mathrm{e}^{-1}\)

One microgram of radioactive sodium \({ }_{11} \mathrm{Na}^{24}\) with a half- life of 15 hours was injected into a living system for a bio assay. How long will it take for the radioactivity to fall to \(25 \%\) of the initial value? (a) 60 hours (b) \(22.5\) hours (c) 375 hours (d) 30 hours

A sample of radioactive substance gave 630 counts per minute and 610 counts per minute at times differing by hour. The decay constant \((\lambda)\) in \(\min ^{-1}\) is given by (a) \(\lambda=\frac{630}{610} \times 60\) (b) \(\mathrm{e}^{60 \mathrm{~A}}=\frac{630}{610}\) (c) \(\lambda=\frac{2.303}{60} \log \frac{610}{630}\) (d) \(\lambda=\frac{2.303}{60} \times \frac{630}{610}\)

What weight of \(\mathrm{C}^{14}\) will have radioactivity one curie if \(\lambda\) (disintegration constant) is \(4.4 \times 10^{-12} \mathrm{sec}^{-1} ?\) (a) \(3.7 \times 10^{-6} \mathrm{~kg}\) (b) \(51 \times 10^{-3} \mathrm{~kg}\) (c) \(1.96 \times 10^{-4} \mathrm{~kg}\) (d) \(1.7 \times 10^{-6} \mathrm{~kg}\)

Match the following: List-I 1\. \(\mathrm{P}^{32}\) 2\. \(\mathrm{Na}^{24}\) 3\. \(\mathrm{Co}^{60}\) 4\. \(\mathrm{I}^{131}\) List-II (i) Location of the tumour in the brain (ii) Location of the blood clots and circulatory disorders (iii) Radiotherapy (iv) Agricultural research

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