A nucleus X, initially at rest, undergoes alpha-decay according to the equation.
92^A X - Z^228 Y +

Q: Why do we use the mass of the proton and neutron to find the binding energy of the products?

The binding energy of any nucleus is the energy required to break it apart into its constituent protons and neutrons. So, we calculate the mass difference between the individual nucleons and the assembled nucleus, and convert that mass defect into energy.

Q: Why is the momentum of the daughter nucleus and alpha particle equal?

The parent nucleus was initially at rest, meaning the initial momentum of the system was zero. To conserve momentum, the vector sum of the momenta of the products must also be zero, which means they must be equal in magnitude and opposite in direction.

Q: Why do we divide by $1.6 \times 10^{-13}$ when calculating the kinetic energy?

The standard SI unit for energy is the Joule. However, in nuclear physics, it is much more convenient to use Mega electron-Volts (MeV). The conversion factor is $1 \text{ MeV} = 1.6 \times 10^{-13} \text{ Joules}$.

Q: Why does the solution use 140 neutrons when calculating the binding energy of the products?

The solution uses a clever shortcut! Instead of calculating the binding energies of $Y$ and $\alpha$ separately, it calculates their sum. The total number of protons in $Y$ and $\alpha$ is $90 + 2 = 92$, and the total number of neutrons is $138 + 2 = 140$. This perfectly matches the nucleon count of the parent nucleus $X$.

Q: Why is the kinetic energy of the daughter nucleus so much smaller than the alpha particle?

Since both particles have the same momentum ($p$), their kinetic energy ($K = p^2 / 2m$) is inversely proportional to their mass. Because the daughter nucleus is much heavier than the alpha particle, it receives a much smaller share of the total kinetic energy.

Question

A nucleus $X$, initially at rest, undergoes alpha-decay according to the equation.
$_{92}^{A}X ightarrow _{Z}^{228}Y + \alpha$
(a) Find the values of $A$ and $Z$ in the above process.
(b) The alpha particle produced in the above process is found to move in a circular track of radius $0.11 	ext{ m}$ in a uniform magnetic field of $3 	ext{ T}$. Find the energy (in MeV) released during the process and the binding energy of the parent nucleus $X$. Given that $m(Y) = 228.03 	ext{ u}$; $m(_{0}^{1}n) = 1.009 	ext{ u}$; $m(_{2}^{4}	ext{He}) = 4.003 	ext{ u}$; $m(_{1}^{1}	ext{H}) = 1.008 	ext{ u}$.

Sigma Percentile · Accepted Answer

### Decoding the Alpha Decay

Let's embark on a fascinating journey into the heart of an atom. We are given a parent nucleus $X$, initially at rest, which undergoes alpha decay to form a daughter nucleus $Y$ and an alpha particle ($\alpha$).

An alpha particle is essentially a helium nucleus, denoted as $_{2}^{4}	ext{He}$. This means it carries a mass number of $4$ and an atomic number of $2$. In any nuclear reaction, the fundamental laws of nature dictate that both the total mass number and the total atomic number must be strictly conserved.

By equating the mass numbers on both sides of our decay equation, we get:

$A = 228 + 4 = 232$

Similarly, equating the atomic numbers yields:

$92 = Z + 2 \implies Z = 90$

This elegantly solves the first part of our problem, revealing the identity of the parent and daughter nuclei.

### The Magnetic Dance of the Alpha Particle

Now, imagine the emitted alpha particle entering a uniform magnetic field. Because it is a charged particle moving perpendicular to the magnetic field, it experiences a Lorentz force that bends its path into a perfect circle. The radius of this circular path is given by the well-known formula:

$r = \frac{mv}{qB}$

We can express the momentum $mv$ in terms of the kinetic energy $K_\alpha$ as $\sqrt{2K_\alpha m_\alpha}$. Substituting this into our radius formula gives:

$r = \frac{\sqrt{2K_\alpha m_\alpha}}{qB}$

Rearranging this to solve for the kinetic energy of the alpha particle, we get:

$K_\alpha = \frac{r^2 q^2 B^2}{2m_\alpha}$

We are given the radius $r = 0.11 	ext{ m}$, the magnetic field $B = 3 	ext{ T}$, and we know the charge of an alpha particle is $q = 2e = 2 	imes 1.6 	imes 10^{-19} 	ext{ C}$. Plugging in these values, along with the mass of the alpha particle in kilograms, and dividing by $1.6 	imes 10^{-13}$ to convert the final answer from Joules to Mega electron-Volts (MeV), we find:

$K_\alpha = 5.21 	ext{ MeV}$

### The Momentum Balance

Since the parent nucleus $X$ was initially at rest, the total initial momentum of the system was zero. To conserve momentum, the daughter nucleus $Y$ and the alpha particle must fly apart with equal and opposite momenta.

$p_Y = p_\alpha \implies \sqrt{2K_Y m_Y} = \sqrt{2K_\alpha m_\alpha}$

This allows us to find the kinetic energy of the recoiling daughter nucleus $Y$:

$K_Y = \left(\frac{m_\alpha}{m_Y}ight) K_\alpha$

Substituting the known masses and the kinetic energy of the alpha particle, we get:

$K_Y = \left(\frac{4.003}{228.03}ight) 	imes 5.21 = 0.09 	ext{ MeV}$

The total energy released in the decay, known as the $Q$-value, is simply the sum of the kinetic energies of the products:

$Q = K_\alpha + K_Y = 5.21 + 0.09 = 5.3 	ext{ MeV}$

### Unveiling the Binding Energy

Finally, we need to determine the binding energy of the parent nucleus $X$. The binding energy of a nucleus is the energy equivalent of its mass defect. A brilliant shortcut here is to realize that the binding energy of the parent nucleus is equal to the total binding energy of the daughter products minus the energy released ($Q$) during the decay.

$BE(X) = BE(	ext{products}) - Q$

To find the binding energy of the products ($Y$ and $\alpha$), we calculate the mass defect between their constituent nucleons and their actual masses. Together, $Y$ and $\alpha$ contain $92$ protons and $140$ neutrons.

$\Delta m = 92 m_p + 140 m_n - m_Y - m_\alpha$

$\Delta m = 92(1.008) + 140(1.009) - 228.03 - 4.003 = 1.963 	ext{ u}$

Converting this mass defect into energy by multiplying by $931.48 	ext{ MeV/u}$ gives:

$BE(	ext{products}) = 1.963 	imes 931.48 = 1828.5 	ext{ MeV}$

Subtracting the $Q$-value from this total gives us the binding energy of the parent nucleus $X$:

$BE(X) = 1828.5 - 5.3 = 1823.2 	ext{ MeV}$

And there we have it! A beautiful interplay of conservation laws, electromagnetism, and mass-energy equivalence.

Visualized Solution

$Alpha Decay Equation$

$Conservation Laws$

$Values of A and Z$

$Radius in Magnetic Field$

$Kinetic Energy of α$

$Substitution$

$Calculating K_{α}$

$Conservation of Momentum$

$Total Energy Released (Q)$

$Binding Energy of X$

$Mass Defect of Products$

$Final Calculation$

The Sigma Insight: Radioactivity

Decoding the Alpha Decay

The Magnetic Dance of the Alpha Particle

The Momentum Balance

Unveiling the Binding Energy

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alpha Decay Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation Laws

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Values of A and Z

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius in Magnetic Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Energy of α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Kα​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Energy Released (Q)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Binding Energy of X

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass Defect of Products

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation