Chapter 11, Problem 3C

To determine

(a)

The number of alpha particles in decay chain starting from U (238) to Pb (206) and U (235) to Pb (207).

Explanation of Solution

Given:

The decay chain of U (238) and U (235) is given as,

$\begin{array}{l}{U}^{238}\to P{b}^{206}\\ U^{235}\to P{b}^{205}\end{array}$.

Calculation:

In the decay chain starting from U (238) to Pb (206).

As we know the mass of alpha particle is 4.

Let there are x number of alpha particles in a decay chain.

$\begin{array}{l}238-4x=206\hfill \\ 32=4x\hfill \\ x=8\hfill \end{array}$

Thus, there are 8 alpha particles.

In the decay chain starting from U (235) to Pb (207).

As we know the mass of alpha particle is 4.

Let there are x number of alpha particles in a decay chain.

$\begin{array}{l}235-4x=207\hfill \\ 28=4x\hfill \\ x=7\hfill \end{array}$

Thus, there are 7 alpha particles.

Conclusion:

In decay chain U (238) to Pb (206) there are 8 alpha particles

In decay chain U (235) to Pb (207) there are 7 alpha particles.

To determine

(b)

The ratio of number of Uranium-238 to lead-206.

Explanation of Solution

Given:

$\begin{array}{l}\text{Number}\text{of}\text{Uranium}\text{initially},N_{0U}=\text{Number}\text{of}\text{Lead}\text{initially},N_{0Pb}\\ \text{time}\text{of}\text{sample},t=4.5\text{billion}\text{years=4}\text{.5}\times {\text{10}}^9\text{years}\end{array}$.

Formula used:

Number of radioactive nuclei at time t is given by,

$\frac{N_t}{N_0}=e^{-\lambda t}$

Where,

$N_t$ = number of radioactive nuclei at time t

$N_0$ = number of radioactive nuclei initially

t = time of sample

$\lambda$ = decay constant.

Calculation:

For U (238), number of Uranium-238 at time t is given by,

$\frac{N_{tU}}{N_{0U}}=e^{-\lambda t}\mathrm{...}(1)$

For Pb (206), number of lead-206 at time t is given by

$\begin{array}{l}\frac{N_{tPb}}{N_{0Pb}}=e^{-\lambda t}\mathrm{...}(2)\\ \frac{N_{tU}}{N_{tPb}}=\frac{e^{-\lambda t}}{e^{-\lambda t}}\\ \frac{N_{tU}}{N_{tPb}}=1\end{array}$.

Conclusion:

The ratio of number of Uranium-238 to lead-206 is, $\frac{N_{tU}}{N_{tPb}}=1$.

To determine

(c)

The ratio of number of Uranium-235 to lead-205.

Explanation of Solution

Given:

$\begin{array}{l}\text{Number}\text{of}\text{Uranium}\text{initially},N_{0U}=\text{Number}\text{of}\text{Lead}\text{initially},N_{0Pb}\\ \text{time}\text{of}\text{sample},t=4.5\text{billion}\text{years=4}\text{.5}\times {\text{10}}^9\text{years}\end{array}$.

Formula used:

Number of radioactive nuclei at time t is given by,

$\frac{N_t}{N_0}=e^{-\lambda t}$

Where,

$N_t$ = number of radioactive nuclei at time t

$N_0$ = number of radioactive nuclei initially

t = time of sample

$\lambda$ = decay constant.

Calculation:

For U (238), Number of radioactive nuclei at time t is given by,

$\frac{N_{tU}}{N_{0U}}=e^{-\lambda t}$

$\begin{array}{l}\frac{N_{tPb}}{N_{0Pb}}=e^{-\lambda t}\\ \frac{N_{tU}}{N_{tPb}}=\frac{e^{-\lambda t}}{e^{-\lambda t}}\\ \frac{N_{tU}}{N_{tPb}}=1\end{array}$.

Conclusion:

The ratio of number of Uranium-235 to lead-207 is $\frac{N_{tU}}{N_{tPb}}=1$.

To determine

(d)

The ratio of number of lead-206 to lead-207.

Explanation of Solution

Formula used:

Number of radioactive nuclei at time t is given by,

$\frac{N_t}{N_0}=e^{-\lambda t}$

Where,

$N_t$ = number of radioactive nuclei at time t

$N_0$ = number of radioactive nuclei initially

t = time of sample

$\lambda$ = decay constant.

Calculation:

Lead-206 and lead-207 are the most stable atoms. Generally, these atoms do not decay at all.

Thus, the ratio of number of Pb (206) nuclei and number of Pb (207) will be equal to the number of atoms of Pb (206) nuclei and number of Pb (207) present initially.

$\frac{N_{t206}}{N_{t207}}=\frac{N_{0206}}{N_{0207}}$.

Conclusion:

The ratio of number of lead-206 to lead-207 is $\frac{N_{t206}}{N_{t207}}=\frac{N_{0206}}{N_{0207}}$.