Radioactivity is the spontaneous disintegration of unstable atomic nuclei accompanied by the emission of ionizing radiation such as alpha particles, beta particles, and gamma rays. The process follows a statistical first-order decay law independent of external physical and chemical conditions.
Basic properties of alpha, beta, and gamma rays, radioactive displacement laws, half-life, mean life, and simple radioactive decay law.
Successive transformations, radioactive equilibrium, secular and transient equilibrium, carbon dating, alpha-decay tunneling theory, and neutrino hypothesis in beta decay.
Conditions: Single radioactive isotope with constant decay constant λ
Result: N(t) = N_0 e^{-\lambda t}
Conditions: Standard exponential decay model
Result: $T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$
| Formula / law | Equation | Variables and units | Conditions |
|---|---|---|---|
| Decay Law Equation | $$N(t) = N_0 e^{-\lambda t}$$ | N(t) = nuclei at time t, $N_0 = initial nuclei$, \lambda = decay constant dimensionless number, s^{-1} | Constant decay probability per unit time |
| Half-Life Formula | $$T_{1/2} = \frac{0.693}{\lambda}$$ | $T_{1/2} = half-life$, \lambda = decay constant s | First-order nuclear kinetics |
| Mean Life Formula | $$\tau = \frac{1}{\lambda} = \frac{T_{1/2}}{0.693}$$ | \tau = mean life, $T_{1/2} = half-life$ s | Valid for exponential decay |
| Activity of a Sample | $$A(t) = \left|\frac{dN}{dt}\right| = \lambda N = A_0 e^{-\lambda t}$$ | A(t) = activity, $A_0 = initial activity$ Becquerel (Bq) or Ci | Standard decay |
| Nuclei Remaining after n Half-Lives | $$N = N_0 \left(\frac{1}{2}\right)^n$$ | $n = number of half-lives elapsed (t / T_{1/2})$ dimensionless | Exact integer or fractional half-lives |
| Secular Equilibrium Condition | $$\lambda_1 N_1 = \lambda_2 N_2$$ | \lambda_1, $N_1 = parent decay constant$ and number; \lambda_2, $N_2 = daughter parameters$ s^{-1} | T_{1/2, 1} \gg T_{1/2, 2} |
| Successive Decay General Equation (Bateman Solution for Daughter) | $$N_2(t) = \frac{\lambda_1 N_{10}}{\lambda_2 - \lambda_1} \left(e^{-\lambda_1 t} - e^{-\lambda_2 t}\right)$$ | $N_2(t) = daughter nuclei$, $N_{10} = initial parent nuclei$ dimensionless | N_{2}(0) = 0 |
| Alpha Decay Energy (Q-value) | $$Q = (m_p - m_d - m_\alpha) c^2$$ | $m_p = parent mass$, $m_d = daughter mass$, $m_\alpha = \alpha particle mass$ J or MeV | Conservation of relativistic energy and momentum |
Draw: A rectangular hyperbola-like decreasing curve of N versus t, showing successive halving of N at equal time intervals T1/2.
Interpret: Demonstrates that the number of radioactive nuclei decreases exponentially with time.
Exam use: Used to graphically determine half-life and decay constant.
Draw: A straight line plot of ln(N) versus t with a negative slope equal to -λ and y-intercept ln(N0).
Interpret: Linearization of the exponential law allows precise calculation of the decay constant from experimental data.
Exam use: Tested in numerical and conceptual questions regarding slope interpretation.
Draw: Count rate versus applied potential difference showing the starting voltage, plateau region, and discharge region.
Interpret: Identifies the working plateau where count rate is independent of voltage fluctuations.
Exam use: Instrumental setup questions in practical-oriented physics examinations.
Answer: 8 g
Answer: Mean life = 20 days, Time taken = 27.72 days
Answer: 0 nuclei
Common error: Radioactivity rate decreases over time because radioactive atoms 'get old' or fatigue.
Correct: The decay probability of an individual nucleus remains constant over time; the decrease in total activity is due to the decreasing number of remaining undecayed nuclei.
Radioactive decay is a random, memoryless Poisson process.
Common error: Gamma rays carry a net positive or negative electric charge.
Correct: Gamma rays are uncharged electromagnetic photons.
Unlike alpha and beta particles, gamma rays are not deflected by electric or magnetic fields.
| Radiation | Nature | Charge | Penetrating Power |
|---|---|---|---|
| Alpha ($\alpha$) | Helium nucleus ($^{4}_{2}\text{He}$) | $+2e$ | Low (Stopped by paper) |
| Beta ($\beta$) | High-speed electron/positron | $-e$ or $+e$ | Moderate (Stopped by thin aluminum sheet) |
| Gamma ($\gamma$) | Electromagnetic quantum | Zero | High (Requires thick lead sheet) |
Choose an option to reveal the explanation.
Explanation: $\alpha$ particles are doubly ionized helium nuclei containing two protons and two neutrons, carrying a charge of $+2e$.
Explanation: $\gamma$ rays are uncharged electromagnetic photons and thus experience no deflection in electric or magnetic fields.
Explanation: According to the radioactive decay law, the rate of disintegration $-\frac{dN}{dt}$ is directly proportional to the number of undecayed nuclei $N$ present at that instant.
Explanation: Using $N(t) = N_0 e^{-\lambda t}$ at $t = T_{1/2}$, $N = N_0/2$, which yields $T_{1/2} = \frac{ln 2}{lambda} approx \frac{0.693}{lambda}.
Explanation: The mean life $\tau$ is defined as the reciprocal of the decay constant, i.e., $\tau = \frac{1}{\lambda}$.
Explanation: The remaining fraction is given by $\left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Explanation: Secular equilibrium occurs when $T_{1/2, 1} gg T_{1/2, 2}$, meaning the parent decays so slowly that its quantity remains practically constant over many daughter half-lives.
Explanation: At secular equilibrium, $\lambda_1 N_1 = \lambda_2 N_2$, which implies that the activity of the parent equals the activity of the daughter ($A_1 = A_2$).
Explanation: Radioactive decay is a nuclear phenomenon and is entirely independent of external physical and chemical conditions such as temperature, pressure, and chemical combination.
Explanation: The SI unit of activity is the Becquerel (Bq), defined as one disintegration per second.
Explanation: In $\beta^-$ decay, a neutron inside the nucleus converts into a proton, an electron ($\beta^-$), and an electron antineutrino.
Explanation: The Q-value represents the energy released, calculated from the mass defect between the parent mass ($m_p$) and the sum of daughter ($m_d$) and $\alpha$ particle ($m_alpha$) masses multiplied by $c^2$.
Explanation: Transient equilibrium takes place when the parent's half-life is longer than the daughter's, but both are of comparable magnitude.
Explanation: Radioactivity is governed by probability laws; while the macroscopic decay rate is predictable, individual nuclear decay is random and memoryless.
Explanation: $T_{1/2} = \frac{0.693}{\lambda} = \frac{0.693}{0.0693\text{ day}^{-1}} = 10\text{ days}$.
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
Explanation: $50%$ decay leaves $50%$ remaining ($1$ half-life). $87.5%$ decay leaves $12.5%$ remaining ($3$ half-lives, since $(1/2)^3 = 1/8 = 12.5%$). The time elapsed between $1$ half-life and $3$ half-lives is $3T_{1/2} - 1T_{1/2} = 2T_{1/2} = 2 \times 20 = 40\text{ minutes}$. Wait, let's re-verify: from $50%$ decayed (remaining $50%$) to $87.5%$ decayed (remaining $12.5%$). Remaining goes from $N_0/2$ to $N_0/8$, which is a drop of two half-lives ($2 \times 20 = 40$ minutes). Let's check options: 40 minutes is option D. Correct index should be 3.
Explanation: Number of half-lives $n = \frac{t}{T_{1/2}} = \frac{6\text{ hours}}{2\text{ hours}} = 3$. Activity $A = A_0 \left(\frac{1}{2}\right)^3 = 64 \times \frac{1}{8} = 8\text{ Bq}$.
Explanation: $N_x = N_0 e^{-3lambda t}$, $N_y = N_0 e^{-\lambda t}$. $\frac{N_x}{N_y} = e^{-2lambda t}$. For $t = \frac{1}{2lambda}$, $\frac{N_x}{N_y} = e^{-2lambda \left(\frac{1}{2lambda}\right)} = e^{-1}$.
Explanation: Ratio of activity $\frac{A}{A_0} = \frac{1250}{10000} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$. Thus, $3$ half-lives elapsed in $30\text{ days}$, so $T_{1/2} = \frac{30}{3} = 10\text{ days}$.
Explanation: At secular equilibrium, $\lambda_1 N_1 = \lambda_2 N_2$, which rearranges to $\frac{N_2}{N_1} = \frac{\lambda_1}{\lambda_2}$.
Explanation: $T_{1/2} = 0.693 \times \tau = 0.693 \times 1440\text{ years} \approx 997.92\text{ years} \approx 998\text{ years}$.
Attempt these questions under a ten-minute limit using a separate timer.
Explanation: Fraction remaining after $t$ days is given by $N/N_0 = e^{-\lambda t}$. Given $e^{-5lambda} = 0.90$. After $20\text{ days}$ ($4$ intervals of $5\text{ days}$), remaining fraction is $(e^{-5lambda})^4 = (0.90)^4 = 0.6561$, which is $65.6%.$
Explanation: In $\beta^+$ decay, a proton converts into a neutron, a positron (positive electron), and an electron neutrino.
Explanation: $T_{1/2} = \frac{0.693}{0.005} = 138.6\text{ days}$.
Explanation: The fraction decayed up to time $t = \tau$ is $1 - \frac{N(\tau)}{N_0} = 1 - e^{-\lambda (1/\lambda)} = 1 - e^{-1} \approx 0.632$ or $63.2%.$
Explanation: Activity is given by $\lambda N$. Since $\lambda = \frac{0.693}{T_{1/2}}$, the decay rates depend on their respective decay constants and remaining numbers of nuclei.
Explanation: $\gamma$ rays are photons emitted during transitions between nuclear energy states; hence, neither $Z$ nor $A$ changes.
Explanation: $\frac{1}{32} = \left(\frac{1}{2}\right)^5$, so $5$ half-lives equal $25\text{ hours}$. Thus, $T_{1/2} = \frac{25}{5} = 5\text{ hours}$.
Explanation: An $\alpha$ particle is $^{4}_{2}\text{He}$, so emission decreases the mass number by $4$ and the atomic number by $2$.
Explanation: The time required for nuclei to reduce to $1/e$ of initial value is the mean life $\tau = \frac{1}{\lambda} = \frac{1}{0.1} = 10\text{ s}$.
Explanation: Wolfgang Pauli proposed the neutrino (antineutrino) hypothesis to explain the continuous energy distribution and conservation of energy/momentum in $\beta$ decay.