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Chapter 13: Problem 12

Find the \(\mathrm{g}\) -value and the kinetic energy of the emitted \(\alpha\) -particle in the \(\alpha\) -decay of (a) \(28 \mathrm{Ra}\) and (b) \({ }_{86}^{220} \mathrm{Rn}\). Given \(m\left({ }_{88}^{226} \mathrm{Ra}\right)=226.02540 \mathrm{u}\) \(m\left(\frac{22}{86} \mathrm{Rn}\right)=222.01750 \mathrm{u}\) \(m\left(_{86}^{22} \mathrm{Rn}\right)=220.01137 \mathrm{u}\) \(m\left({ }_{84}^{216} \mathrm{Po}\right)=216.00189 \mathrm{u}\)

Short Answer

Expert verified
Ra decay: \( g = 4.94 \text{ MeV} \); Rn decay: \( g = 6.41 \text{ MeV} \).

Step by step solution

01

Understand the Problem

We need to find the \( g \)-value and the kinetic energy of the emitted \( \alpha \)-particle for both \( ^{226}_{88}\textrm{Ra} \) and \( ^{220}_{86}\textrm{Rn} \). This means calculating the energy difference between the parent and daughter nuclei when an \( \alpha \)-particle (\( ^4_2\textrm{He} \)) is emitted.
02

Calculate the Mass Defect for (a) \( ^{226}_{88}\textrm{Ra} \)

The mass defect \( \Delta m \) is given by the difference in mass before and after the \( \alpha \)-decay:\[ \Delta m = m(^{226}_{88}\textrm{Ra}) - \left[m(^{222}_{86}\textrm{Rn}) + m(^{4}_{2}\textrm{He})\right] \]Given \( m(^{226}_{88}\textrm{Ra}) = 226.02540 \), \( m(^{222}_{86}\textrm{Rn}) = 222.01750 \), and \( m(^{4}_{2}\textrm{He}) = 4.00260 \) amu (standard mass for \( \alpha\)-particles).Calculate the mass defect \( \Delta m = 226.02540 - (222.01750 + 4.00260) = 0.00530 \) amu.
03

Convert Mass Defect to Energy for (a) \( ^{226}_{88}\textrm{Ra} \)

Use the energy-mass equivalence \( E = \Delta m \cdot c^2 \), where \( c^2 \approx 931.5 \text{ MeV/amu} \). Thus,\[ E = 0.00530 \cdot 931.5 \approx 4.94 \text{ MeV} \]
04

Calculate the Mass Defect for (b) \( ^{220}_{86}\textrm{Rn} \)

The mass defect \( \Delta m \) is given by the difference in mass before and after the \( \alpha \)-decay:\[ \Delta m = m(^{220}_{86}\textrm{Rn}) - \left[m(^{216}_{84}\textrm{Po}) + m(^{4}_{2}\textrm{He})\right] \]Given \( m(^{220}_{86}\textrm{Rn}) = 220.01137 \), \( m(^{216}_{84}\textrm{Po}) = 216.00189 \), and \( m(^{4}_{2}\textrm{He}) = 4.00260 \) amu.Calculate the mass defect \( \Delta m = 220.01137 - (216.00189 + 4.00260) = 0.00688 \) amu.
05

Convert Mass Defect to Energy for (b) \( ^{220}_{86}\textrm{Rn} \)

Use the energy-mass equivalence \( E = \Delta m \cdot c^2 \), where \( c^2 \approx 931.5 \text{ MeV/amu} \). Thus,\[ E = 0.00688 \cdot 931.5 \approx 6.41 \text{ MeV} \]
06

Find the g-value and Kinetic Energy

The \( g \)-value (\( Q \)-value) of the reaction essentially indicates the total energy released and is equivalent to the kinetic energy of the \( \alpha \)-particle in these cases because the recoil energy of the daughter nucleus is negligible.

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Key Concepts

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

Mass Defect
In nuclear physics, the mass defect is a crucial concept when studying processes like alpha decay. It refers to the difference in mass between a composite nucleus and the sum of the masses of its individual components.
  • This difference arises because some of the mass has been converted into binding energy, which holds the nucleus together.
  • During an alpha decay, a nucleus emits an alpha particle 鈥� a helium nucleus 鈥� and transforms into a lighter element.
  • The mass defect can be calculated by taking the initial mass of the parent nucleus and subtracting the total mass of the daughter nucleus and the emitted alpha particle.
Understanding mass defect helps us calculate the energy released during the decay process, as explained by Einstein's famous equation, \( E = \Delta m \cdot c^2 \).
Q-value
The Q-value in nuclear reactions represents the total energy released or absorbed during a process. In the case of alpha decay, it shows the energy released when the unstable nucleus releases an alpha particle.
  • The Q-value is crucial because it determines the energy with which the alpha particle is emitted.
  • In essence, it equals the energy equivalent of the mass defect calculated during the alpha decay.
  • Moreover, because the recoil of the daughter nucleus is negligible, the Q-value can be considered approximately equal to the kinetic energy of the emitted alpha particle.
Thus, the Q-value not only reveals the energy dynamics of the decay but also the speed and energy of the alpha particle subsequently released.
Kinetic Energy
Kinetic energy in the context of alpha decay refers to the energy carried away by the emitted alpha particle. Since the daughter nucleus maintains most of its energy, the kinetic energy is almost equivalent to the Q-value of the decay process.
  • Kinetic energy accounts for the motion of the alpha particle after it is dislodged from the parent nucleus.
  • It is calculated using the relation \( E = \Delta m \cdot c^2 \), helping to determine how energetic the particle motion will be.
  • This energy measurement is critical in nuclear physics, providing insights into both nuclear structure and stability.
An understanding of kinetic energy is essential to comprehend the outcomes of nuclear reactions such as alpha decay.
Nuclear Physics
Nuclear physics is the field that studies the constituents and interactions of atomic nuclei. It plays a significant role in understanding the processes like alpha decay.
  • Nuclear physics explores how different forces and particles interact within the nucleus, including the attractive strong nuclear force and the repulsive electrostatic force.
  • Alpha decay, a type of radioactive decay, is an area of particular interest, involving the emission of an alpha particle from a nucleus.
  • This field also examines the energy changes occurring during nuclear reactions, using concepts such as mass defect, binding energy, and Q-value.
By delving into nuclear physics, scientists gain insight into the stability of elements and the sources of immense energy in nuclear reactions.

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Most popular questions from this chapter

Obtain the binding energy of the nuclei \({ }_{25}^{56} \mathrm{Fe}\) and \({ }_{83}^{209} \mathrm{Bi}\) in units of \(\mathrm{MeV}\) from the following data: \(m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.934939 \mathrm{u} \quad m\left({ }_{83}^{209} \mathrm{Bi}\right)=208.980388 \mathrm{u}\)

A radioactive isotope has a half-life of \(T\) years. How long will it take the activity to reduce to a) \(3.125 \%\), b) \(1 \%\) of its original value?

Suppose India had a target of producing by \(2020 \mathrm{AD}, 200,000 \mathrm{MW}\) of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was \(25 \%\). How much amount of fissionable uranium would our country need per year by \(2020 ?\) Take the heat energy per fission of \({ }^{235} \mathrm{U}\) to be about \(200 \mathrm{MeV}\).

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei \({ }_{20}^{41} \mathrm{Ca}\) and \({ }_{13}^{27} \mathrm{Al}\) from the following data: \(m\left({ }_{20}^{40} \mathrm{Ca}\right)=39.962591 \mathrm{u}\) \(m\left({ }_{20}^{41} \mathrm{Ca}\right)=40.962278 \mathrm{u}\) \(m\left({ }_{13}^{26} \mathrm{Al}\right)=25.986895 \mathrm{u}\) \(m\left({ }_{13}^{27} \mathrm{Al}\right)=26.981541 \mathrm{u}\)

A given coin has a mass of \(3.0\) g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of \({ }_{29}^{63} \mathrm{Cu}\) atoms (of mass \(\left.62.92960 \mathrm{u}\right)\).
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