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Modern Physics
Elementary Theory of Alpha Decay
Alpha decay is a radioactive process in which an unstable atomic nucleus ejects an alpha particle, transforming into a daughter nucleus with a mass number smaller by 4 and an atomic number smaller by 2. The elementary quantum-mechanical theory explains this process through quantum tunneling of the pre-formed alpha particle through a Coulomb potential barrier.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of alpha decay, conservation laws of mass number and atomic number, Geiger-Nuttall empirical law relating decay constant and range of alpha particles, and basic energetics (Q-value).
PGT extension
Quantum mechanical tunneling model (Gamow theory), barrier penetration probability, derivation of decay constant, and dependence on disintegration energy.
Prerequisites
Conservation of energy and linear momentum in nuclear reactions
De Broglie wavelength and elementary wave mechanics
Coulomb's law and potential energy of point charges
Energetics of Alpha Decay
The Q-value of alpha decay is defined as $Q = (m_P - m_D - m_alpha) c^{2}$, representing the total energy released during the transformation.
The disintegration energy Q is shared between the recoiling daughter nucleus and the alpha particle inversely proportional to their masses to conserve linear momentum.
For an alpha decay to occur spontaneously, the Q-value must be strictly positive (Q > 0).
The Coulomb Potential Barrier
Outside the nuclear radius R, the potential energy experienced by the alpha particle is purely electrostatic: V(r) = (1 / 4πε_0) (2 Z e²) / r.
At the nuclear surface, the attractive nuclear forces form a potential well, while the repulsive electrostatic force forms a barrier of height V_max exceeding the kinetic energy of the alpha particle.
Classically, an alpha particle with kinetic energy less than the barrier height cannot escape the nucleus.
Quantum Mechanical Tunneling (Gamow Theory)
According to quantum mechanics, particles possess a non-zero probability wave that penetrates classically forbidden regions.
George Gamow treated the alpha particle as an independent entity bouncing back and forth inside the nuclear potential well, striking the barrier f times per second.
The decay constant λ is the product of the assault frequency f and the quantum transmission coefficient T ($\lambda = f T)$.
Geiger-Nuttall Law
Geiger and Nuttall established an empirical linear relation between the logarithm of the decay constant (λ) and the logarithm of the range (R) of the emitted alpha particle in air: log $\lambda = A \log R + B$.
Higher energy alpha particles have shorter lifetimes (larger decay constants) because they encounter a thinner and lower effective potential barrier.
The law is naturally explained by Gamow's tunneling theory.
Derivations
Transmission Coefficient for a Coulomb Barrier
Conditions: Assuming a rectangular barrier approximation or WKB approximation for the Coulomb potential V(r) = (2Ze²/(4πε_0 r))
Apply the WKB approximation to find the decay constant or barrier penetration factor $P = exp(-2 \int _{R}^{r_0} k(r) dr)$, where k(r) = &sqrt;(2μ(V(r) - E)) / ℏ.
Evaluate the integral between the nuclear radius R and the classical turning point r_0 where V($r_0) = E$.
Perform the integration analytically by substituting $r = r_0 \cos^{2}\theta$ to obtain the Gamow factor.
Express the transmission coefficient P in terms of the charge numbers, fine structure constant, and velocity of the alpha particle.
Result: $P = exp(-2 G)$, where $G = ( \pi Z_1 Z_2 e^{2} / 4\pi\epsilon_0 \hbar v ) - 4 ( Z_1 Z_2 R / 4\pi\epsilon_0 \hbar^2 \mu )^(1/2)$
Kinetic Energy Distribution of Decay Products
Conditions: Rest frame of the parent nucleus, conservation of total energy and linear momentum.
Set initial momentum of parent nucleus to zero: $p_P = 0$.
Z = atomic number of parent nucleus, e = elementary charge, r = separation distance Joules
r > R
Barrier Penetration Probability (Approximate)
$$T \approx \exp \left( -2 \frac{e^2}{\hbar \nu} Z \right)$$
$\nu = velocity of \alpha particle$, Z = atomic number, e = electronic charge Dimensionless
Low energy limit
Daughter Kinetic Energy
$$K_D = \frac{m_\alpha}{m_\alpha + m_D} Q$$
m_$\alpha = mass of \alpha particle$, $m_D = mass of daughter$, $Q = Q-value$ Joules or MeV
Momentum conservation
Graphs and experiments
Nuclear Potential Well and Coulomb Barrier
Draw: Potential energy V(r) on y-axis versus radial distance r on x-axis, showing a deep attractive well inside nuclear radius R, a Coulomb barrier peak of height V_max, and a horizontal line representing alpha particle energy E.
Interpret: Illustrates that the alpha particle energy E is less than the barrier peak V_max, making classical escape impossible, yet quantum tunneling allows escape.
Exam use: Frequently asked in conceptual multiple-choice questions to explain why alpha particles emerge with energy less than the barrier height.
Geiger-Nuttall Plot
Draw: Straight lines plotting log(λ) against log(R) for different radioactive series (Uranium, Thorium, Actinium).
Interpret: Shows parallel straight lines confirming that isotopes with higher energy (longer range R) have exponentially shorter half-lives.
Exam use: Tested via interpretation of slopes and intercepts or matching decay constants with particle ranges.
Cloud Chamber Track of Alpha Particles
Draw: Straight, dense, parallel ionization tracks of uniform length originating from a point source.
Interpret: Demonstrates that monoenergetic alpha particles travel a well-defined range in matter before losing all their energy.
Exam use: Used in experimental physics questions related to the detection and characteristic properties of alpha radiation.
Solved numericals
TGT core
Calculate the Q-value of the alpha decay of 226Ra to 222Rn, given the masses: m(^226Ra) = 226.0254 u, m(^222Rn) = 222.0175 u, and m(α) = 4.0015 u. (1 $u = 931.5 MeV/c^2)$
Write the mass defect Δ$m = m(Parent) - m(Daughter) - m(\alpha)$.
Substitute the given atomic masses: Δ$m = 226.0254 u - 222.0175 u - 4.0015 u$.
Calculate Δ$m = 226.0254 - 226.0190 = 0.0064 u$.
Multiply by 931.5 MeV/u to find Q: $Q = 0.0064 \times 931.5 MeV$.
Answer: 5.96 MeV
PGT extension
An alpha particle is emitted with a kinetic energy of 4.8 MeV from a nucleus with atomic number Z = 90. Assuming the daughter nucleus has mass number A = 228, find the kinetic energy of the recoiling daughter nucleus.
Identify the given values: $K_alpha = 4.8 MeV$, m_alpha ≈ 4 u, m_D ≈ A - $4 = 228 - 4 = 224 u$.
Use momentum conservation relation between kinetic energies: K_D / $K_alpha = m_alpha / m_D$.
Substitute the mass values: $K_D = 4.8 MeV \times (4 / 224)$.
Evaluate the division: 4 / $224 = 1 / 56$, so $K_D = 4.8 / 56 MeV$.
Answer: 0.0857 MeV (or 85.7 keV)
PGT extension
If the disintegration energy Q for an alpha decay is 5.4 MeV, determine the exact kinetic energy of the emitted alpha particle, taking the exact mass of the alpha particle as 4.0015 u and the daughter nucleus mass as 218 u.
Recall the formula relating alpha kinetic energy to Q: $K_alpha = Q [ m_D / (m_alpha + m_D) ]$.
Substitute the given mass values: $m_D = 218 u$, $m_alpha = 4.0015 u$, total mass = 222.0015 u.
Calculate the mass ratio: 218 / 222.0015 = 0.98197.
The alpha particle is assumed to pre-form inside the nucleus prior to emission, though nuclear structure models view it as a cluster formation probability.
The WKB approximation used in calculating the barrier penetration assumes the potential varies slowly compared to the de Broglie wavelength of the alpha particle.
Common misconceptions
Common error: Alpha particles possess kinetic energy greater than the height of the Coulomb barrier inside the nucleus.
Correct: The kinetic energy of the alpha particle is less than the barrier peak, and it escapes via quantum mechanical tunneling.
Classically, an energy deficit would prevent escape; quantum mechanics permits barrier penetration.
Common error: All alpha particles emitted from a specific radioactive isotope have continuous kinetic energy spectra like beta decay.
Correct: Alpha particles from a given transition are monoenergetic.
Two-body decay kinematics dictates a fixed energy sharing between the alpha particle and the daughter nucleus.
Quick revision
Alpha decay reduces mass number by 4 and atomic number by 2.
Spontaneous alpha decay requires a positive Q-value.
Gamow theory successfully explains alpha decay via quantum tunneling through the Coulomb barrier.
Geiger-Nuttall law states a linear relation between log(λ) and log(R).
Must remember
The Q-value of alpha decay must be strictly positive ($Q > 0$) for the decay to occur spontaneously.
Alpha particles are monoenergetic because alpha decay is a two-body decay process governed by momentum and energy conservation.
Classically, an alpha particle cannot escape the nucleus because its kinetic energy is less than the Coulomb barrier height.
Gamow's theory explains alpha decay using quantum mechanical tunneling through the Coulomb barrier.
The Geiger-Nuttall law establishes a linear relationship between $\log \lambda$ and $\log R$, where $R$ is the range of the alpha particle in air.
Glossary
Q-valueThe total disintegration energy released during alpha decay, given by $Q = (m_P - m_D - m_\alpha) c^2$.
Gamow Factor ($G$)A dimensionless integral parameter in the WKB approximation representing the attenuation of the wave function across the barrier.
Assault Frequency ($f$)The frequency with which the alpha particle strikes the inside of the nuclear potential barrier per second.
Transmission Coefficient ($T$)The probability that an alpha particle successfully tunnels through the potential barrier during a single assault.
Geiger-Nuttall LawAn empirical formula $\log \lambda = A \log R + B$ connecting the decay constant and the range of alpha particles in air.
Alpha Decay Kinematics
Parameter
Alpha Particle ($m_\alpha$)
Daughter Nucleus ($M_D$)
Kinetic Energy Share
$K_\alpha = \frac{M_D}{M_P} Q$
$K_D = \frac{m_\alpha}{M_P} Q$
Momentum Magnitude
$p = \sqrt{2 m_\alpha K_\alpha}$
$p = \sqrt{2 M_D K_D}$ (equal and opposite)
Practice Quiz
Choose an option to reveal the explanation.
1. What is the primary condition for an $\alpha$ decay to occur spontaneously?
Explanation: For an $\alpha$ decay to occur spontaneously, the total disintegration energy must be strictly positive ($Q > 0$), meaning the parent mass exceeds the sum of the daughter and $\alpha$ particle masses.
2. Are $\alpha$ particles emitted from a given radioactive isotope monoenergetic or continuous in energy?
Explanation: $\alpha$ particles from a specific transition are monoenergetic because $\alpha$ decay is a two-body decay process, resulting in a fixed energy sharing between the $\alpha$ particle and the daughter nucleus.
3. How is the kinetic energy of the emitted $\alpha$ particle ($K_\alpha$) expressed in terms of the Q-value and masses?
Explanation: From momentum conservation, the kinetic energy of the $\alpha$ particle is shared inversely proportional to masses, yielding $K_\alpha = \frac{M_D}{M_P} Q$ where $M_P = M_D + m_\alpha$.
4. What physical mechanism allows an $\alpha$ particle to escape the nucleus even when its kinetic energy is less than the Coulomb barrier height?
Explanation: According to quantum mechanics, particles have a non-zero probability wave that penetrates classically forbidden regions, a phenomenon known as quantum mechanical tunneling (Gamow theory).
5. What does the Geiger-Nuttall law relate?
Explanation: The Geiger-Nuttall law establishes an empirical linear relation between the logarithm of the decay constant ($\lambda$) and the logarithm of the range ($R$) of the $\alpha$ particle in air: $\log \lambda = A \log R + B$.
6. Why do higher energy $\alpha$ particles generally exhibit shorter lifetimes (larger decay constants)?
Explanation: Higher energy $\alpha$ particles penetrate closer to the classical turning point, effectively encountering a thinner and lower potential barrier, which significantly increases their transmission coefficient and decay constant.
7. According to Gamow's theory, the decay constant $\lambda$ is expressed as the product of the assault frequency $f$ and which parameter?
Explanation: Gamow treated the $\alpha$ particle as bouncing back and forth inside the nuclear potential well, striking the barrier $f$ times per second, so the decay constant is given by $\lambda = f T$.
8. What is the nature of the potential energy function $V(r)$ experienced by an $\alpha$ particle outside the nuclear radius $R$?
Explanation: Outside the nuclear radius $R$, the attractive nuclear forces become negligible, and the potential energy experienced by the $\alpha$ particle is purely electrostatic: $V(r) = \frac{1}{4\pi\epsilon_0} \frac{2(Z-2)e^2}{r}$.
9. If the parent nucleus has mass $M_P$, daughter nucleus $M_D$, and $\alpha$ particle mass $m_\alpha$, what is the expression for the Q-value?
Explanation: The Q-value represents the mass energy converted into kinetic energy: $Q = [m(Parent) - m(Daughter) - m(\alpha)] c^2$.
10. Which of the following describes the kinetic energy of the daughter nucleus $K_D$ in $\alpha$ decay?
Explanation: By energy and momentum conservation, the daughter nucleus receives a kinetic energy inversely proportional to its mass share: $K_D = \frac{m_\alpha}{M_P} Q$.
11. In the context of the Gamow factor $G$, what does $r_0$ represent?
Explanation: In the Gamow factor integral $G = \frac{\sqrt{2m}}{\hbar} \int_{R}^{r_0} \sqrt{V(r) - E} \, dr$, $R$ is the nuclear radius and $r_0$ is the classical turning point where $V(r_0) = E$.
12. What misconception is corrected by Gamow's theory of $\alpha$ decay?
Explanation: Classically, it was believed $\alpha$ particles needed kinetic energy greater than the barrier height to escape; Gamow corrected this by showing they escape via quantum mechanical tunneling despite having lower energy.
13. Which parameter in the Geiger-Nuttall law ($\log \lambda = A \log R + B$) represents the slope?
Explanation: The equation is in the linear form $y = m x + c$, where $A$ is the slope and $B$ is the intercept.
14. What conservation law ensures that the $\alpha$ particle and daughter nucleus recoil in opposite directions with specific energy ratios?
Explanation: Linear momentum conservation requires that the momenta of the $\alpha$ particle and the recoiling daughter nucleus are equal in magnitude and opposite in direction ($p_\alpha = p_D$).शैक्षणिक
15. What is the units of the transmission coefficient $T$ in Gamow's decay theory?
Explanation: The transmission coefficient $T$ is a probability ratio representing the fraction of wave function transmitted through the barrier, making it dimensionless.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. In an $\alpha$ decay process, if the Q-value is $5.5\text{ MeV}$, what is the approximate kinetic energy of the emitted $\alpha$ particle from a parent nucleus with mass number $A = 212$?
Explanation: Using $K_\alpha \approx \frac{A - 4}{A} Q = \frac{208}{212} \times 5.5\text{ MeV} \approx 5.39\text{ MeV} \approx 5.4\text{ MeV}$ using mass number proportionality.
2. According to the Geiger-Nuttall law, if the range of an $\alpha$ particle in air is increased, how does its decay constant $\lambda$ behave?
Explanation: Since $\log \lambda = A \log R + B$ with positive slope $A$, a larger range $R$ corresponds to higher energy and a larger decay constant $\lambda$ (shorter half-life).
3. What is the physical significance of the classical turning point $r_0$ in the Gamow barrier penetration integral?
Explanation: The classical turning point $r_0$ is where the kinetic energy $E$ equals the potential energy $V(r_0)$, marking the boundary of the classically forbidden region.
4. If the nuclear radius $R$ of a parent nucleus decreases while the Q-value remains constant, how is the transmission coefficient $T$ affected?
Explanation: A smaller nuclear radius increases the width of the Coulomb barrier that the $\alpha$ particle must tunnel through, increasing the Gamow factor $G$ and decreasing the transmission coefficient $T$.
5. In the disintegration energy equation $Q = [m(Parent) - m(Daughter) - m(\alpha)] c^2$, what does a negative Q-value signify?
Explanation: A negative Q-value means mass-energy must be supplied from the outside for the reaction to take place; hence, spontaneous $\alpha$ decay ($Q > 0$) cannot occur.
6. Why are $\alpha$ particles monoenergetic whereas $\beta$ particles exhibit a continuous energy spectrum?
Explanation: $\alpha$ decay is a two-body decay, leading to a fixed energy sharing dictated by momentum conservation. $\beta$ decay involves three bodies (nucleus, electron/positron, and neutrino/antineutrino), distributing energy continuously.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Calculate the kinetic energy of an $\alpha$ particle emitted in a decay with a Q-value of $4.0\text{ MeV}$ from a parent nucleus of mass number $A = 236$.
2. Which of the following correctly represents the dependence of the Gamow factor $G$ on particle velocity $\nu$ in low-energy barrier penetration?
Explanation: The approximate transmission probability formula $T \approx \exp(-2 \frac{e^2}{\hbar \nu} Z)$ shows that the exponent and attenuation factor depend inversely on velocity $\nu$.
3. What is the primary reason why heavy nuclei undergo $\alpha$ decay rather than proton emission?
Explanation: The $\alpha$ particle possesses a very high binding energy ($28.3\text{ MeV}$), making its formation inside heavy nuclei energetically favorable ($Q > 0$), whereas single proton emission usually has a negative Q-value.
4. If the decay constant of a radioactive substance emitting $\alpha$ particles is $\lambda = 1.5 \times 10^{-5} \text{ s}^{-1}$ and the assault frequency is $f = 10^{21} \text{ s}^{-1}$, what is the approximate transmission coefficient $T$?
Explanation: Since $\lambda = f T$, we have $T = \frac{\lambda}{f} = \frac{1.5 \times 10^{-5}}{10^{21}} = 1.5 \times 10^{-26}$.
5. In the Coulomb potential barrier formula $V(r) = \frac{1}{4\pi\epsilon_0} \frac{2(Z-2)e^2}{r}$, what does the term $2(Z-2)$ represent?
Explanation: The numerator represents the electrostatic interaction between the $\alpha$ particle ($2e$) and the daughter nucleus of atomic number $Z' = Z-2$ (charge $(Z-2)e$).
6. How does an increase in the atomic number $Z$ of the parent nucleus affect the height of the Coulomb potential barrier?
Explanation: The Coulomb potential $V(r)$ is directly proportional to the atomic number $Z$ of the parent nucleus, so a higher $Z$ raises the electrostatic barrier peak.
7. Which of the following statements about Gamow's theory is INCORRECT?
Explanation: Gamow's theory specifically accounts for $\alpha$ particles having kinetic energy *less* than the barrier peak, overcoming it via tunneling.
8. If the kinetic energy of the daughter nucleus is $K_D$ and the $\alpha$ particle is $K_\alpha$, what is the ratio $\frac{K_D}{K_\alpha}$?
Explanation: Since $K_\alpha = \frac{M_D}{M_P} Q$ and $K_D = \frac{m_\alpha}{M_P} Q$, the ratio is $\frac{K_D}{K_\alpha} = \frac{m_\alpha}{M_D}$.
9. What physical quantity does the area under the curve $\sqrt{V(r) - E}$ between $R$ and $r_0$ determine in $\alpha$ decay?
Explanation: The integral $\int_{R}^{r_0} \sqrt{V(r) - E} \, dr$ forms the core of the exponent in the Gamow factor $G$, which governs barrier transparency.
10. Why is $\alpha$ decay rarely observed in light nuclei ($A < 150$)?
Explanation: In light nuclei, the mass of the parent is generally less than or equal to the sum of the daughter and $\alpha$ masses ($Q \le 0$), making spontaneous $\alpha$ decay energetically forbidden.