Question

For a radionuclide that obeys Equation $32-3$,

(a) find the percentage of the nuclide remaining after a time equal to $1/\lambda$.

(b) What time in units of $1/\lambda$would be required for $90\%$of the nuclide to decay?

(c) What time in units of $1/\lambda$would be required for$99\%$ of the nuclide to decay?

Short answer

(a) The percentage of the nuclide remaining after a time equal to $1/\lambda$is $36.8\%$.

(b) Time in units of $1/\lambda$that would be required for 90% of the nuclide to decay is $2.3025/\lambda$.

(c) Time in units of $1/\lambda$that would be required for 99% of the nuclide to decay is$4.6051/\lambda$.

Step-by-step solution

Part (a) Step 1: Given information

The radionuclide obeys the equation$\ln \left(\frac{N}{N_0}\right)=-\lambda t$.

Time$=t=1/\lambda$

Part (a) Step 2: Calculation

By substituting the given value of time in the equation, we get,

$$\begin{gathered} \ln \left(\frac{N}{N_0}\right)=-\lambda \times \frac{1}{\lambda } \\ \ln \left(\frac{N}{N_0}\right)=-1 \\ \frac{N}{N_0}=e^{-1} \\ \frac{N}{N_0}=0.368=36.8\% \end{gathered}$$

Part (a) Step 3: Final answer

Hence, it can be calculated that $36.8\%$nuclide is remaining as per the given conditions.

Part (b) Step 1: Given information

The radionuclide obeys the equation $\ln \left(\frac{N}{N_0}\right)=-\lambda t$.

The nuclide decays $90\%$.

Part (b) Step 2: Calculation

Since the nuclide decays $90\%$, the remaining amount would be $10\%$.

Hence,

$\frac{N}{N_0}=\frac{10}{100}=0.1$

By substituting this value in the given equation, we get,

$$\begin{gathered} \ln \left(0.1\right)=-\lambda t \\ -2.3025=-\lambda t \\ t=\frac{2.3025}{\lambda } \end{gathered}$$

Part (b) Step 3: Final answer

The time required is $2.3025/\lambda$for $90\%$of the nuclide to decay.

Part (c) Step 1: Given information

The radionuclide obeys the equation $\ln \left(\frac{N}{N_0}\right)=-\lambda t$.

The nuclide decays $99\%$.

Part (c) Step 2: Calculation

Since the nuclide decays $99\%$, the remaining amount would be $1\%$.

Hence,

$\frac{N}{N_0}=\frac{1}{100}=0.01$

By substituting this value in the given equation, we get,

$$\begin{gathered} \ln \left(0.01\right)=-\lambda t \\ -4.6051=-\lambda t \\ t=\frac{4.6051}{\lambda } \end{gathered}$$

Part (c) Step 3: Final answer

The time required is $4.6051/\lambda$ for$99\%$ of the nuclide to decay.