Question:
The half-life of a radioactive substance is 20 minutes. $\frac{1}{3}$rd part of substance has decayed in time $t_1$ and $\frac{2}{3}$rd part of it has decayed in time $t_2$. Then, ($t_2 - t_1$) is nearly
The half-life of a radioactive substance is 20 minutes. $\frac{1}{3}$rd part of substance has decayed in time $t_1$ and $\frac{2}{3}$rd part of it has decayed in time $t_2$. Then, ($t_2 - t_1$) is nearly
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For any first-order process, the time taken to decay from fraction $f_1$ to $f_2$ depends only on the ratio of the remaining quantities.
Since the ratio of the remaining amounts $\frac{N(t_1)}{N(t_2)} = \frac{2/3}{1/3} = 2$, this decay corresponds exactly to one half-life.
Since the ratio of the remaining amounts $\frac{N(t_1)}{N(t_2)} = \frac{2/3}{1/3} = 2$, this decay corresponds exactly to one half-life.
Updated On: Jul 22, 2026
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Solution and Explanation
Step 1: Understanding the Question:
The question is based on radioactive decay kinetics.
We need to determine the time difference $(t_2 - t_1)$ between two points of decay of a radioactive substance.
Step 2: Key Formula or Approach:
Radioactive decay is a first-order kinetics process. The amount of radioactive substance remaining is given by:
\[ N(t) = N_0 e^{-\lambda t} \] This can be rewritten in terms of time as:
\[ t = \frac{1}{\lambda} \ln \left( \frac{N_0}{N(t)} \right) \] where $N_0$ is the initial quantity and $N(t)$ is the remaining quantity at time $t$.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]
Step 3: Detailed Explanation:
• Let us analyze the amount of substance remaining at times $t_1$ and $t_2$.
• At time $t_1$, $\frac{1}{3}\text{rd}$ of the substance has decayed. Therefore, the remaining fraction is:
\[ N(t_1) = N_0 \left(1 - \frac{1}{3}\right) = \frac{2}{3} N_0 \]
• The time $t_1$ is given by:
\[ t_1 = \frac{1}{\lambda} \ln \left( \frac{N_0}{\frac{2}{3} N_0} \right) = \frac{1}{\lambda} \ln \left( \frac{3}{2} \right) \]
• At time $t_2$, $\frac{2}{3}\text{rd}$ of the substance has decayed. Therefore, the remaining fraction is:
\[ N(t_2) = N_0 \left(1 - \frac{2}{3}\right) = \frac{1}{3} N_0 \]
• The time $t_2$ is given by:
\[ t_2 = \frac{1}{\lambda} \ln \left( \frac{N_0}{\frac{1}{3} N_0} \right) = \frac{1}{\lambda} \ln(3) \]
• Now, we calculate the time difference $t_2 - t_1$:
\[ t_2 - t_1 = \frac{1}{\lambda} \ln(3) - \frac{1}{\lambda} \ln\left(\frac{3}{2}\right) \] \[ t_2 - t_1 = \frac{1}{\lambda} \ln\left( \frac{3}{\frac{3}{2}} \right) = \frac{1}{\lambda} \ln(2) \]
• Substituting $\lambda = \frac{\ln 2}{T_{1/2}}$:
\[ t_2 - t_1 = \frac{\ln 2}{\frac{\ln 2}{T_{1/2}}} = T_{1/2} \]
• Since the half-life is 20 minutes, we have:
\[ t_2 - t_1 = 20\text{ minutes} \]
Step 4: Final Answer:
The value of $(t_2 - t_1)$ is equal to the half-life of the substance, which is 20 minutes.
The question is based on radioactive decay kinetics.
We need to determine the time difference $(t_2 - t_1)$ between two points of decay of a radioactive substance.
Step 2: Key Formula or Approach:
Radioactive decay is a first-order kinetics process. The amount of radioactive substance remaining is given by:
\[ N(t) = N_0 e^{-\lambda t} \] This can be rewritten in terms of time as:
\[ t = \frac{1}{\lambda} \ln \left( \frac{N_0}{N(t)} \right) \] where $N_0$ is the initial quantity and $N(t)$ is the remaining quantity at time $t$.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]
Step 3: Detailed Explanation:
• Let us analyze the amount of substance remaining at times $t_1$ and $t_2$.
• At time $t_1$, $\frac{1}{3}\text{rd}$ of the substance has decayed. Therefore, the remaining fraction is:
\[ N(t_1) = N_0 \left(1 - \frac{1}{3}\right) = \frac{2}{3} N_0 \]
• The time $t_1$ is given by:
\[ t_1 = \frac{1}{\lambda} \ln \left( \frac{N_0}{\frac{2}{3} N_0} \right) = \frac{1}{\lambda} \ln \left( \frac{3}{2} \right) \]
• At time $t_2$, $\frac{2}{3}\text{rd}$ of the substance has decayed. Therefore, the remaining fraction is:
\[ N(t_2) = N_0 \left(1 - \frac{2}{3}\right) = \frac{1}{3} N_0 \]
• The time $t_2$ is given by:
\[ t_2 = \frac{1}{\lambda} \ln \left( \frac{N_0}{\frac{1}{3} N_0} \right) = \frac{1}{\lambda} \ln(3) \]
• Now, we calculate the time difference $t_2 - t_1$:
\[ t_2 - t_1 = \frac{1}{\lambda} \ln(3) - \frac{1}{\lambda} \ln\left(\frac{3}{2}\right) \] \[ t_2 - t_1 = \frac{1}{\lambda} \ln\left( \frac{3}{\frac{3}{2}} \right) = \frac{1}{\lambda} \ln(2) \]
• Substituting $\lambda = \frac{\ln 2}{T_{1/2}}$:
\[ t_2 - t_1 = \frac{\ln 2}{\frac{\ln 2}{T_{1/2}}} = T_{1/2} \]
• Since the half-life is 20 minutes, we have:
\[ t_2 - t_1 = 20\text{ minutes} \]
Step 4: Final Answer:
The value of $(t_2 - t_1)$ is equal to the half-life of the substance, which is 20 minutes.
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