Problem 89
Question
Boron-10 emits alpha particles and cesium-137 emits beta particles. Write balanced nuclear reactions for each radioactive decay.
Step-by-Step Solution
Verified Answer
The balanced nuclear equations for the radioactive decay of Boron-10 and Cesium-137 are:
Boron-10 (Alpha decay):
\[
_5^{10}\text{B} \rightarrow _2^4\text{He} + _3^6\text{Li}
\]
Cesium-137 (Beta decay):
\[
_{55}^{137}\text{Cs} \rightarrow _{-1}^0\text{e} + _{56}^{137}\text{Ba}
\]
1Step 1: Boron-10 decay (Alpha decay)
1. We are given that Boron-10 (\(_5^{10}\)B) emits an alpha particle (\(_2^4\)He) in its decay. To find the resulting nucleus, we need to subtract the alpha particle from the original nucleus:
\[
_5^{10}\text{B} \rightarrow _2^4\text{He} + _{?}^?\text{X}
\]
2. Balance the reaction by subtracting the mass numbers (top numbers) and atomic numbers (bottom numbers) to find the resulting nucleus:
\[
_3^6\text{Li} = _5^{10}\text{B} - _2^4\text{He}
\]
3. Write the balanced nuclear equation:
\[
_5^{10}\text{B} \rightarrow _2^4\text{He} + _3^6\text{Li}
\]
The balanced nuclear equation for the alpha decay of Boron-10 is:
\[
_5^{10}\text{B} \rightarrow _2^4\text{He} + _3^6\text{Li}
\]
2Step 2: Cesium-137 decay (Beta decay)
1. We are given that Cesium-137 (\(_{55}^{137}\)Cs) emits a beta particle (\(_{-1}^0\)e) in its decay. To find the resulting nucleus, we need to balance the original nucleus with the beta particle:
\[
_{55}^{137}\text{Cs} \rightarrow _{-1}^0\text{e} + _{?}^?\text{X}
\]
2. Balance the reaction by adding the atomic numbers (bottom numbers) of the nuclide and beta particle, while keeping the mass number (top number) the same, to find the resulting nucleus:
\[
_{56}^{137}\text{Ba} = _{55}^{137}\text{Cs} + _{-1}^0\text{e}
\]
3. Write the balanced nuclear equation:
\[
_{55}^{137}\text{Cs} \rightarrow _{-1}^0\text{e} + _{56}^{137}\text{Ba}
\]
The balanced nuclear equation for the beta decay of Cesium-137 is:
\[
_{55}^{137}\text{Cs} \rightarrow _{-1}^0\text{e} + _{56}^{137}\text{Ba}
\]
Key Concepts
Alpha DecayBeta DecayBalanced Nuclear Equations
Alpha Decay
Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle. An alpha particle consists of two protons and two neutrons, effectively making it equivalent to a helium nucleus, usually denoted as \(_2^4\text{He}\). This leads to a decrease in both the atomic number and the mass number of the original nucleus by two and four, respectively.
For example, in the case of Boron-10 (\(_5^{10}\text{B}\)), the nucleus emits an alpha particle and transforms into Lithium-6 (\(_3^6\text{Li}\)). We can see this in the balanced nuclear equation:
For example, in the case of Boron-10 (\(_5^{10}\text{B}\)), the nucleus emits an alpha particle and transforms into Lithium-6 (\(_3^6\text{Li}\)). We can see this in the balanced nuclear equation:
- \(_5^{10}\text{B} \rightarrow _2^4\text{He} + _3^6\text{Li}\)
Beta Decay
Beta decay occurs when a neutron in an atomic nucleus is transformed into a proton, while emitting a beta particle, which is an electron denoted as \(_{-1}^0\text{e}\). In this process, the atomic number of the element increases by one, while the mass number remains unchanged. This means the element transforms into a different element that is one place higher in the periodic table.
Consider the example of Cesium-137 (\(_{55}^{137}\text{Cs}\)). When it undergoes beta decay, it emits a beta particle and becomes Barium-137 (\(_{56}^{137}\text{Ba}\)). The balanced nuclear equation is:
Consider the example of Cesium-137 (\(_{55}^{137}\text{Cs}\)). When it undergoes beta decay, it emits a beta particle and becomes Barium-137 (\(_{56}^{137}\text{Ba}\)). The balanced nuclear equation is:
- \(_{55}^{137}\text{Cs} \rightarrow _{-1}^0\text{e} + _{56}^{137}\text{Ba}\)
Balanced Nuclear Equations
Balanced nuclear equations are essential for understanding nuclear reactions like alpha and beta decay. These equations ensure that both the mass numbers and atomic numbers are conserved before and after the reaction.
In alpha decay, the total atomic number decreases by two and the total mass number decreases by four. For beta decay, the atomic number increases by one whereas the mass number remains constant. By following these principles, we can accurately write and balance nuclear equations.
Understanding these balances helps in predicting products of nuclear reactions and ensures that we follow the law of conservation of mass and charge. This is crucial in fields that work with radioactivity, such as nuclear physics, medicine, and energy.
In alpha decay, the total atomic number decreases by two and the total mass number decreases by four. For beta decay, the atomic number increases by one whereas the mass number remains constant. By following these principles, we can accurately write and balance nuclear equations.
Understanding these balances helps in predicting products of nuclear reactions and ensures that we follow the law of conservation of mass and charge. This is crucial in fields that work with radioactivity, such as nuclear physics, medicine, and energy.
Other exercises in this chapter
Problem 87
Explain how energy loss and nuclear stability are related to radioactive decay.
View solution Problem 88
Explain what must occur before a radioactive atom stops to undergo further radioactive decay.
View solution Problem 90
Determine what was wrong with Dalton’s theory and provide the most recent version of the atomic structure.
View solution Problem 91
Cathode-Ray Tube Describe a cathode-ray tube and how it operates.
View solution