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Chapter 21: Problem 15

Decay of which nucleus will lead to the following products: (a) uranium-235 by alpha decay; (b) aluminium-26 by positron emission; \((\mathbf{c})\) deuterium by alpha decay; (d) yttrium-90 by beta decay?

Short Answer

Expert verified
The parent nuclei for each decay process are: a) Plutonium-238 (\(_{94}^{238}Pu\)) for Uranium-235 by alpha decay, b) Magnesium-26 (\(_{12}^{26}Mg\)) for Aluminium-26 by positron emission, c) No possible parent nucleus for deuterium by alpha decay since deuterium cannot be a product of alpha decay, d) Strontium-90 (\(_{38}^{90}Sr\)) for Yttrium-90 by beta decay.

Step by step solution

01

a) Uranium-235 by Alpha decay.

In alpha decay, the parent nucleus emits an alpha particle, which is a helium-4 nucleus (2 protons and 2 neutrons). The general decay equation is: \(A_{p}\rightarrow A_{d}+_{2}^{4}He\) Where \(A_{p}\) is the parent nucleus, and \(A_{d}\) is the daughter nucleus. For the given decay: \(_{92}^{238}X\rightarrow _{92}^{235}U +_{2}^{4}He\) Thus, the parent nucleus is Plutonium-238, \(_{94}^{238}Pu\).
02

b) Aluminium-26 by Positron Emission

In positron emission, the parent nucleus decays by converting a proton into a neutron and emitting a positron, which is represented by \(e^{+}\). The general decay equation is: \(A_{p}\rightarrow A_{d} + e^{+}\) For the given decay: \(_{13}^{26}X\rightarrow _{12}^{26}Al +_{1}^{0}e^{+}\) Thus, the parent nucleus is Magnesium-26, \(_{12}^{26}Mg\).
03

c) Deuterium by Alpha decay

In an alpha decay, the parent nucleus emits an alpha particle, which is a helium-4 nucleus (2 protons and 2 neutrons). The general decay equation is: \(A_{p}\rightarrow A_{d} + _{2}^{4}He\) For the given decay: \(_{z}^{A}X\rightarrow _{1}^{2}H + _{2}^{4}He\) In this case, we see that deuterium has been mentioned as a product instead of a parent nucleus. Deuterium cannot be a product of an alpha decay since the decaying nucleus must have larger mass and atomic numbers than the product nucleus. Therefore, there is no possible parent nucleus for this decay process.
04

d) Yttrium-90 by Beta decay

In a beta decay, the parent nucleus decays by converting a neutron into a proton and emitting a beta particle, which is an electron, represented by \(e^{-}\). The general decay equation is: \(A_{p}\rightarrow A_{d} + e^{-}\) For the given decay: \(_{38}^{90}X\rightarrow _{39}^{90}Y + _{-1}^{0}e^{-}\) Thus, the parent nucleus is Strontium-90, \(_{38}^{90}Sr\).

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Most popular questions from this chapter

Charcoal samples from Stonehenge in England were burned in \(\mathrm{O}_{2},\) and the resultant \(\mathrm{CO}_{2}\) gas bubbled into a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (limewater), resulting in the precipitation of \(\mathrm{CaCO}_{3}\). The \(\mathrm{CaCO}_{3}\) was removed by filtration and dried. A 788 -mg sample of the \(\mathrm{CaCO}_{3}\) had a radioactivity of $1.5 \times 10^{-2}$ Bq due to carbon-14. By comparison, living organisms undergo 15.3 disintegrations per minute per gram of carbon. Using the half-life of carbon-14, 5700 yr, calculate the age of the charcoal sample.

Assume that Bismuth- 213 decays to a stable nucleus by a series of two alpha and two beta emissions. What is the stable nucleus that is formed?

What particle is produced during the following decay processes: (a) actinium-215 decays to francium-211; (b) boron- 13 decays to carbon-13; (c) holmium-151 decays to terbium- \(147 ;\) (d) carbon-11 decays to boron-11?

Each of the following nuclei undergoes either beta decay or positron emission. Predict the type of emission for each: (a) \({ }_{38}^{90} \mathrm{Sr}\) (b) \({ }_{38}^{85} \mathrm{Sr}\) (d) sulfur-30. (c) potassium- 40 ,

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