Radioactivity and Nuclear Reaction

a) Definitions

• Decay constant ($\lambda$): the ratio of the rate of decay to the number of undecayed nuclei present at that instant, $\lambda = -\dfrac{dN/dt}{N}$. It is the probability per unit time that a nucleus decays. Also, from $N = N_0 e^{-\lambda t}$, it is the reciprocal of the time in which $N$ falls to $1/e$ (about 37%) of $N_0$. Unit: s$^{-1}$.
• Half-life ($T_{1/2}$): the time in which half of the radioactive nuclei in a sample decay. $T_{1/2} = \dfrac{0.693}{\lambda}$.

b) Derivation of $N = N_0 e^{-\lambda t}$

Setup: Let $N_0$ be the number of radioactive nuclei at time $t = 0$ and $N$ the number left undecayed at time $t$. In a short time $dt$, let $dN$ nuclei decay.

1. By the radioactive decay law, the rate of decay is proportional to the number of nuclei present:

$$-\frac{dN}{dt} \propto N \quad\Rightarrow\quad \frac{dN}{dt} = -\lambda N$$

where $\lambda$ is the decay constant. The negative sign shows that $N$ decreases with time. 2. Separate the variables:

$$\frac{dN}{N} = -\lambda\, dt$$

3. Integrate:

$$\int \frac{dN}{N} = -\lambda \int dt \quad\Rightarrow\quad \ln N = -\lambda t + C$$

4. At $t = 0$, $N = N_0$, so $C = \ln N_0$:

$$\ln N - \ln N_0 = -\lambda t \quad\Rightarrow\quad \ln\frac{N}{N_0} = -\lambda t$$

5. Take the exponential of both sides:

$$\boxed{N = N_0\, e^{-\lambda t}}$$

So the number of undecayed nuclei decreases exponentially with time.

(a) Radioactivity

Radioactivity is the spontaneous disintegration of the nuclei of heavy, unstable atoms (such as uranium and radium), with the emission of $\alpha$-particles, $\beta$-particles and $\gamma$-rays, without any external cause.

(b) Derivation of $N = N_0 e^{-\lambda t}$

Setup: Let $N_0$ be the number of radioactive nuclei at time $t = 0$ and $N$ the number left undecayed at time $t$. In a short time $dt$, let $dN$ nuclei decay.

1. By the radioactive decay law, the rate of decay is proportional to the number of nuclei present:

$$-\frac{dN}{dt} \propto N \quad\Rightarrow\quad \frac{dN}{dt} = -\lambda N$$

where $\lambda$ is the decay constant. The negative sign shows that $N$ decreases with time. 2. Separate the variables:

$$\frac{dN}{N} = -\lambda\, dt$$

3. Integrate:

$$\int \frac{dN}{N} = -\lambda \int dt \quad\Rightarrow\quad \ln N = -\lambda t + C$$

4. At $t = 0$, $N = N_0$, so $C = \ln N_0$:

$$\ln N - \ln N_0 = -\lambda t \quad\Rightarrow\quad \ln\frac{N}{N_0} = -\lambda t$$

5. Take the exponential of both sides:

$$\boxed{N = N_0\, e^{-\lambda t}}$$

So the number of undecayed nuclei decreases exponentially with time.

(c) Significance of the decay curve

   N
   ^
N0 +*
   | *
   |  *
N0/2 + - - *
   |       : *
N0/4 + - - - - - -*
   |       :      :  *
   |       :      :      *  .  .  .  .  .  .  (never reaches zero)
   +-------+------+------------------------------> t
   0       T     2T

The graph of $N$ against $t$ is an exponential curve. It approaches the time axis but never touches it, which means that in theory a radioactive sample takes an infinite time to decay completely. So we cannot state the total life of a sample; instead we use the half-life $T_{1/2}$ or the mean life $\tau = 1/\lambda$. A flatter curve (smaller $\lambda$) means a longer-lived isotope.

(d) Time for 25% of radium to remain

Given: $T_{1/2} = 1620$ years, $\dfrac{N}{N_0} = 25\% = \dfrac{1}{4}$.

Using $\dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^{t/T_{1/2}}$:

$$\frac{1}{4} = \left(\frac{1}{2}\right)^{t/T_{1/2}} = \left(\frac{1}{2}\right)^2 \quad\Rightarrow\quad \frac{t}{T_{1/2}} = 2$$ $$t = 2 \times 1620 = \boxed{3240\ \text{years}}$$

(a) Radioactive decay laws (Rutherford and Soddy)

1. Radioactive decay is spontaneous and random. It is not affected by external conditions such as temperature, pressure or chemical combination.
2. In each decay a nucleus emits an $\alpha$-particle or a $\beta$-particle (often with $\gamma$-rays) and changes into the nucleus of a new element.
3. The rate of decay at any instant is directly proportional to the number of undecayed nuclei present at that instant:

$$-\frac{dN}{dt} \propto N \quad\text{or}\quad \frac{dN}{dt} = -\lambda N$$

(b) Derivation of $N = N_0 e^{-\lambda t}$

Setup: Let $N_0$ be the number of radioactive nuclei at time $t = 0$ and $N$ the number left undecayed at time $t$. In a short time $dt$, let $dN$ nuclei decay.

1. By the radioactive decay law, the rate of decay is proportional to the number of nuclei present:

$$-\frac{dN}{dt} \propto N \quad\Rightarrow\quad \frac{dN}{dt} = -\lambda N$$

where $\lambda$ is the decay constant. The negative sign shows that $N$ decreases with time. 2. Separate the variables:

$$\frac{dN}{N} = -\lambda\, dt$$

3. Integrate:

$$\int \frac{dN}{N} = -\lambda \int dt \quad\Rightarrow\quad \ln N = -\lambda t + C$$

4. At $t = 0$, $N = N_0$, so $C = \ln N_0$:

$$\ln N - \ln N_0 = -\lambda t \quad\Rightarrow\quad \ln\frac{N}{N_0} = -\lambda t$$

5. Take the exponential of both sides:

$$\boxed{N = N_0\, e^{-\lambda t}}$$

So the number of undecayed nuclei decreases exponentially with time.

(c) (i) Half-life

The sample is left with 50% of its original nuclei in 2 years. By definition, this time is the half-life:

$$\boxed{T_{1/2} = 2\ \text{years}}$$

(c) (ii) Initial number of nuclei

Given: $N = 10^6$ after $t = 8$ years, $T_{1/2} = 2$ years.

Number of half-lives: $n = \dfrac{t}{T_{1/2}} = \dfrac{8}{2} = 4$

$$N = N_0\left(\frac{1}{2}\right)^n \quad\Rightarrow\quad N_0 = N \times 2^n = 10^6 \times 2^4 = 10^6 \times 16$$ $$\boxed{N_0 = 1.6 \times 10^7\ \text{nuclei}}$$

a) Radioactivity

Radioactivity is the spontaneous disintegration of unstable atomic nuclei with the emission of $\alpha$-particles, $\beta$-particles and $\gamma$-rays.

b) Comparing rates of disintegration after 18 hours

Given: $N_{0A} = N_{0B} = N_0$, $T_A = 3$ h, $T_B = 9$ h, $t = 18$ h.

Number of half-lives: $n_A = \dfrac{18}{3} = 6$, $n_B = \dfrac{18}{9} = 2$

Nuclei left:

$$N_A = N_0\left(\tfrac{1}{2}\right)^6 = \frac{N_0}{64}, \qquad N_B = N_0\left(\tfrac{1}{2}\right)^2 = \frac{N_0}{4}$$

Rate of disintegration (activity) $R = \lambda N$, with $\lambda = \dfrac{0.693}{T_{1/2}}$:

$$\frac{R_A}{R_B} = \frac{\lambda_A N_A}{\lambda_B N_B} = \frac{T_B}{T_A}\cdot\frac{N_A}{N_B} = \frac{9}{3}\times\frac{N_0/64}{N_0/4} = 3 \times \frac{4}{64} = \frac{3}{16}$$ $$\boxed{R_A : R_B = 3 : 16}$$

So after 18 hours, sample B disintegrates faster than A (about 5.3 times faster), even though A has the shorter half-life.

c) Relation between half-life and decay constant

From the decay law $N = N_0 e^{-\lambda t}$:

At $t = T_{1/2}$ (half-life), $N = \dfrac{N_0}{2}$. Putting this in $N = N_0 e^{-\lambda t}$:

$$\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \quad\Rightarrow\quad e^{\lambda T_{1/2}} = 2 \quad\Rightarrow\quad \lambda T_{1/2} = \ln 2$$ $$\boxed{T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}}$$

d) Percentage decayed after five half-lives

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^5 = \frac{1}{32} = 0.03125 = 3.125\%$$

Percentage decayed $= 100\% - 3.125\% = \boxed{96.875\% \approx 96.9\%}$

a) Definitions

• Half-life ($T_{1/2}$): the time in which half of the radioactive nuclei present in a sample decay. $T_{1/2} = \dfrac{0.693}{\lambda}$.
• Mean life ($\tau$): the average time for which a nucleus of the sample survives before it decays. It equals the reciprocal of the decay constant: $\tau = \dfrac{1}{\lambda}$.

b) Mean life of the sample

Given: $T_{1/2} = 50$ years. (The mass, 2 g, does not affect the mean life, because half-life and mean life do not depend on the amount of substance.)

$$\lambda = \frac{0.693}{T_{1/2}}, \qquad \tau = \frac{1}{\lambda} = \frac{T_{1/2}}{0.693} = 1.443\, T_{1/2}$$ $$\tau = \frac{50}{0.693} \approx \boxed{72.1\ \text{years}}$$

c) Value of $T_3$

$N_0/2$, $N_0/4$ and $N_0/8$ are reached after 1, 2 and 3 half-lives. So

$$T_3 = 3 \times T_{1/2} = 3 \times 10 = \boxed{30\ \text{years}}$$

a) Definitions

• Half-life ($T_{1/2}$): the time in which half of the radioactive nuclei present in a sample decay.
• Mean life ($\tau$): the average lifetime of all the nuclei of a radioactive sample before they decay; $\tau = \dfrac{1}{\lambda} = 1.443\,T_{1/2}$.

b) Half-life and decay constant from fig (a)

From the graph, the number of nuclei falls from 1000 to 500 in 1 day (and from 500 to 250 in the next day). So

$$\boxed{T_{1/2} = 1\ \text{day}}$$ $$\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1\ \text{day}} = 0.693\ \text{day}^{-1}$$

In SI units: $\lambda = \dfrac{0.693}{24 \times 3600\ \text{s}} \approx \boxed{8.02 \times 10^{-6}\ \text{s}^{-1}}$

c) Curve with the larger decay constant in fig (b)

Curve 1 has the greatest decay constant. Activity $A = A_0 e^{-\lambda t}$, so the larger $\lambda$ is, the faster the activity falls. Curve 1 falls most steeply and reaches $A_0/2$ in the shortest time (smallest half-life, and $\lambda = 0.693/T_{1/2}$ is largest).

(a) Comparison of $\beta$-particles and $\gamma$-rays

(b) Decay constant

The decay constant is the reciprocal of the time in which the number of undecayed nuclei falls to $1/e$ of its initial value. Equivalently, it is the probability of decay of a nucleus per unit time, $\lambda = \dfrac{-dN/dt}{N}$.

(c) Activity

$$T_{1/2} = 5700\ \text{y} = 5700\times3.156\times10^{7} \approx 1.80\times10^{11}\ \text{s}$$ $$\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.80\times10^{11}} = 3.85\times10^{-12}\ \text{s}^{-1}$$ $$A = \lambda N = 3.85\times10^{-12}\times10^{22} = 3.85\times10^{10}\ \text{Bq}$$

(a) One curie is the activity of a radioactive sample in which $3.7\times10^{10}$ disintegrations occur per second.

(b) From the decay law $N = N_0e^{-\lambda t}$, after one half-life $t = T_{1/2}$ the number of nuclei is $N_0/2$:

$\frac{N_0}{2} = N_0e^{-\lambda T_{1/2}} \;\Rightarrow\; e^{\lambda T_{1/2}} = 2 \;\Rightarrow\; \lambda T_{1/2} = \ln 2$

$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$

(c) The number of half-lives is $n = 25/10 = 2.5$.

$\frac{N}{N_0} = \left(\frac12\right)^{2.5} = 0.177$

About 17.7 % of the sample remains.

(a) Natural radioactivity is the spontaneous emission of radiation ($\alpha$, $\beta$ or $\gamma$) by naturally occurring unstable nuclei, without any external influence.

(b) The rate of decay is proportional to the number of undecayed nuclei $N$ present at that instant:

$\frac{dN}{dt} = -\lambda N$

where $\lambda$ is the decay constant and the negative sign shows that $N$ decreases. Rearranging and integrating from $N_0$ at $t=0$ to $N$ at time $t$:

$\int_{N_0}^{N}\frac{dN}{N} = -\lambda\int_0^t dt \;\Rightarrow\; \ln\frac{N}{N_0} = -\lambda t$

$\boxed{N = N_0e^{-\lambda t}}$

(c) Current sensitivity of a galvanometer is the deflection produced per unit current: $I_s = \theta/I = NBA/C$. Voltage sensitivity is the deflection produced per unit voltage: $V_s = \theta/V = NBA/(CR)$, where $R$ is the resistance of the galvanometer.

(d) The number of half-lives is $n = 38/3.8 = 10$.

$m = m_0\left(\frac12\right)^{10} = \frac{1024}{1024} = 1\ \text{mg}$

So 1 mg of radon is left.

(B) 6%

After 4 half-lives, the fraction left is $\left(\tfrac{1}{2}\right)^4 = \tfrac{1}{16} = 6.25\%$, which is about 6%.

(a) Derivation of $N = N_0 e^{-\lambda t}$

Setup: Let $N_0$ be the number of radioactive nuclei at time $t = 0$ and $N$ the number left undecayed at time $t$. In a short time $dt$, let $dN$ nuclei decay.

1. By the radioactive decay law, the rate of decay is proportional to the number of nuclei present:

$$-\frac{dN}{dt} \propto N \quad\Rightarrow\quad \frac{dN}{dt} = -\lambda N$$

where $\lambda$ is the decay constant. The negative sign shows that $N$ decreases with time. 2. Separate the variables:

$$\frac{dN}{N} = -\lambda\, dt$$

3. Integrate:

$$\int \frac{dN}{N} = -\lambda \int dt \quad\Rightarrow\quad \ln N = -\lambda t + C$$

4. At $t = 0$, $N = N_0$, so $C = \ln N_0$:

$$\ln N - \ln N_0 = -\lambda t \quad\Rightarrow\quad \ln\frac{N}{N_0} = -\lambda t$$

5. Take the exponential of both sides:

$$\boxed{N = N_0\, e^{-\lambda t}}$$

So the number of undecayed nuclei decreases exponentially with time.

(b) (i) Why the data are scattered

Radioactive decay is a random and spontaneous process. We cannot predict exactly when a particular nucleus will decay, so the number of decays counted in each interval fluctuates about the average value. Background radiation adds further random counts. So the points lie scattered about a smooth curve.

(b) (ii) Half-life

From the best-fit curve, the count rate falls from about 50 cps at $t = 0$ to about 25 cps at $t \approx 50$ s, and from 25 cps to about 12.5 cps at $t \approx 100$ s. Equal time is taken for each halving, so

$$\boxed{T_{1/2} \approx 50\ \text{s}}$$

(b) (iii) Sample with a larger decay constant

The curve will fall more steeply (it drops to low count rates more quickly). Activity $A = A_0 e^{-\lambda t}$, so a larger $\lambda$ makes the activity decrease faster, and the half-life $T_{1/2} = 0.693/\lambda$ becomes shorter. The curve is still exponential, but it reaches half of its starting value in less than 50 s.