Write equations for the following nuclear reactions. (a) bombardment of \(^{7} \mathrm{Li}\) with protons to produce \(^{8} \mathrm{Be}\) and \(\gamma\) rays (b) bombardment of \(^{9} \mathrm{B}\) with \(_{1}^{2} \mathrm{H}\) to produce \(^{10} \mathrm{B}\) (c) bombardment of \(^{14} \mathrm{N}\) with neutrons to produce \(^{14} \mathrm{C}\)

Short Answer

Expert verified
The nuclear reactions can be written as follows: (a) \(^{7} \mathrm{Li} + ^{1} \mathrm{H} \rightarrow ^{8} \mathrm{Be} + \gamma\) (b) \(^{9} \mathrm{B} + _{1}^{2} \mathrm{H} \rightarrow ^{10} \mathrm{B}\) (c) \(^{14} \mathrm{N} + _{0}^{1} \mathrm{n} \rightarrow ^{14} \mathrm{C}\)

Step by step solution

01

Write the nuclear reaction for \(^{7} \mathrm{Li}\) and protons

In the first scenario, \(^{7} \mathrm{Li}\) is bombarded with protons (\(^{1} \mathrm{H}\)). The result is an \(^{8} \mathrm{Be}\) atom and \(\gamma\) rays. The equation for this nuclear reaction can be written as follows: \(^{7} \mathrm{Li} + ^{1} \mathrm{H} \rightarrow ^{8} \mathrm{Be} + \gamma\)
02

Write the nuclear reaction for \(^{9} \mathrm{B}\) and \(_{1}^{2} \mathrm{H}\)

In the second scenario, \(^{9} \mathrm{B}\) is bombarded with \(_{1}^{2} \mathrm{H}\). The result is an \(^{10} \mathrm{B}\) atom. The equation for this nuclear reaction can be written as follows: \(^{9} \mathrm{B} + _{1}^{2} \mathrm{H} \rightarrow ^{10} \mathrm{B}\)
03

Write the nuclear reaction for \(^{14} \mathrm{N}\) and neutrons

In the third scenario, \(^{14} \mathrm{N}\) is bombarded with neutrons (\(_{0}^{1} \mathrm{n}\)). The result is an \(^{14} \mathrm{C}\) atom. The equation for this nuclear reaction can be written as follows: \(^{14} \mathrm{N} + _{0}^{1} \mathrm{n} \rightarrow ^{14} \mathrm{C}\)

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Key Concepts

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

Nuclear Reaction Equations
Nuclear reaction equations are a way of representing the processes occurring in nuclear reactions, where the nuclei of atoms interact to form different elements. These equations follow the principle of conservation of mass and energy, ensuring that the number of protons and neutrons is the same on both sides of the equation.

For example, when writing a nuclear reaction equation, each reactant and product is represented by its atomic symbol, mass number, and atomic number. In the exercise provided, the nuclear reaction equation for the bombardment of lithium-7 with protons is represented as: \[ ^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow ^{8}_{4}\text{Be} + \text{ }\text{gamma rays (}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }) \].

This shows that a lithium-7 nucleus and a proton combine to form a beryllium-8 nucleus while emitting gamma rays, fulfilling the criteria for a balanced nuclear equation.
Isotope Bombardment
Isotope bombardment is a technique used in nuclear chemistry to induce a nuclear reaction by firing a particle at an isotope. This can result in a change in the target's atomic nucleus, creating a new isotope or even a new element.

In the exercise examples, various isotopes are bombarded with different particles. Such as protons (\(^{1}_{1}\text{H}\)), deuterons (\(_{1}^{2}\text{H}\)), and neutrons (\(_{0}^{1}\text{n}\)). The result of each bombardment is the production of a new isotope or element, showcasing the fundamental concept of nuclear transmutation. Isotopes are chosen based on their potential to create desired reactions, as well as considerations like stability and decay products.
Nuclear Transmutation
Nuclear transmutation is the process of converting one element or isotope into another through nuclear reaction or radioactive decay. It is an essential process in nuclear chemistry and plays a vital role in various applications, including the generation of energy in nuclear reactors and the creation of new elements.

In the context of the provided exercise, transmutation occurs when an isotope absorbs a particle and changes into another isotope or element, such as nitrogen-14 absorbing a neutron to become carbon-14. This change is represented by the reaction equation: \[ ^{14}_{7}\text{N} + _{0}^{1}\text{n} \rightarrow ^{14}_{6}\text{C} \].

Understanding nuclear transmutation is crucial for managing nuclear waste, synthesizing new materials, and studying elemental formation in the universe. It represents the transformative power of nuclear physics to alter the very identity of elements at the atomic level.

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Most popular questions from this chapter

Just as the uranium series is called the "4n \(+2^{\prime \prime}\) series, the thorium series can be called the "4n" series and the actinium series the "4n \(+3 "\) series. A \(4 n+1 "\) series has also been established, with \(^{241} \mathrm{Pu}\) as the parent nuclide. To which series does each of the following belong: (a) \(\frac{214}{83} \mathrm{Bi} ;\) (b) \(\frac{216}{84} \mathrm{Po} ;\) (c) \(\frac{215}{85} \mathrm{At}\) (d) \(\frac{235}{92} \mathrm{U} ?\)

The disintegration rate for a sample containing \(_{27}^{60} \mathrm{Co}\) as the only radioactive nuclide is 6740 dis \(\mathrm{h}^{-1}\). The half-life of 20 Co is 5.2 years. Estimate the number of atoms of \(_{27}^{60}\) Co in the sample.

A sample containing \(_{88}^{224} \mathrm{Ra},\) which decays by \(\alpha\) -particle emission, disintegrates at the following rate, expressed as disintegrations per minute or counts per minute \((\mathrm{cpm}): t=0,1000 \mathrm{cpm} ; t=1 \mathrm{h}\) \(992 \mathrm{cpm} ; t=10 \mathrm{h}, 924 \mathrm{cpm} ; t=100 \mathrm{h}, 452 \mathrm{cpm}\) \(t=250 \mathrm{h}, 138 \mathrm{cpm} .\) What is the half-life of this nuclide?

The carbon-14 dating method is based on the assumption that the rate of production of \(^{14} \mathrm{C}\) by cosmic ray bombardment has remained constant for thousands of years and that the ratio of \(^{14} \mathrm{C}\) to \(^{12} \mathrm{C}\) has also remained constant. Can you think of any effects of human activities that could invalidate this assumption in the future?

A lunar rock was analyzed for argon by mass spectrometry and for potassium by atomic absorption. The results of these analyses showed that the sample contained \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon and \(0.083 \%\) of potassium. The half-life of potassium- 40 is \(1.248 \times\) \(10^{9} \mathrm{y} \cdot\) Calculate the age of the lunar rock.

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