Question

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as : (where $\lambda$ is the decay constant)

• A. $\frac{\ln(3/2)}{\lambda}$
• B. $\frac{1}{2} \frac{\ln 2}{\lambda}$
• C. $\frac{2 \ln 2}{\lambda}$
• D. $\frac{\ln 2}{\lambda}$

Hint:

Remember that the decay equation relates the amount remaining. Always subtract the decayed amount from the initial amount before using the formula.
Use log properties: \(\ln a - \ln b = \ln(a/b)\).

Answer 1

The Correct Option is A

To solve this problem, we need to determine the time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed, using the decay constant \lambda. This involves understanding radioactive decay and the mathematical model that describes it.

The number of undecayed nuclei at any time t is given by the equation:

N(t) = N_0 e^{-\lambda t}

where:

• N(t) is the number of undecayed nuclei at time t
• N_0 is the initial number of nuclei
• \lambda is the decay constant

First, calculate the time t_1 when a quarter of the nuclei have decayed. This means three-quarters are left, so:

N(t_1) = \frac{3}{4}N_0

Substitute in the decay formula:

\frac{3}{4}N_0 = N_0 e^{-\lambda t_1}

Cancel N_0 on both sides:

\frac{3}{4} = e^{-\lambda t_1}

Take the natural logarithm of both sides to solve for t_1:

-\lambda t_1 = \ln\left(\frac{3}{4}\right)

t_1 = -\frac{1}{\lambda} \ln\left(\frac{3}{4}\right)

Next, calculate the time t_2 when half of the nuclei have decayed. This means half are left, so:

N(t_2) = \frac{1}{2}N_0

Substitute in the decay formula:

\frac{1}{2}N_0 = N_0 e^{-\lambda t_2}

Cancel N_0 on both sides:

\frac{1}{2} = e^{-\lambda t_2}

Take the natural logarithm of both sides to solve for t_2:

-\lambda t_2 = \ln\left(\frac{1}{2}\right)

t_2 = -\frac{1}{\lambda} \ln\left(\frac{1}{2}\right)

Now, find the time gap \Delta t = t_2 - t_1:

\Delta t = -\frac{1}{\lambda} \ln\left(\frac{1}{2}\right) - \left(-\frac{1}{\lambda} \ln\left(\frac{3}{4}\right)\right)

\Delta t = -\frac{1}{\lambda} (\ln\left(\frac{1}{2}\right) - \ln\left(\frac{3}{4}\right))

Applying the logarithmic identity \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right):

\Delta t = -\frac{1}{\lambda} \ln\left(\frac{\frac{1}{2}}{\frac{3}{4}}\right)

\Delta t = -\frac{1}{\lambda} \ln\left(\frac{1}{2} \times \frac{4}{3}\right)

\Delta t = -\frac{1}{\lambda} \ln\left(\frac{2}{3}\right)

\Delta t = \frac{1}{\lambda} \ln\left(\frac{3}{2}\right)

Thus, the time gap is \frac{\ln(3/2)}{\lambda}, which matches the given correct answer.