Question

A radioactive nucleus $A$ with a half-life $T$, decays into a nucleus $B$. At $t = 0$, there is no nucleus $B$. At sometime t, the ratio of the number of $B$ to that of $A$ is $0.3$. Then, $t$ is given by :

• A. $t = \frac{T}{2} \frac{\log \, 2}{\log \, (1.3)}$
• B. $t = T \frac{\log \, (1.3)}{\log \, 2}$
• C. $t = T \, \log (1.3)$
• D. $t = \frac{T }{\log \, (1.3)}$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the time $t$ at which the ratio of the number of nuclei $B$ to nuclei $A$ is $0.3$, given that the half-life of $A$ is $T$. We will use the concept of radioactive decay, where nuclei $A$ decay into $B$.

Radioactive decay follows the equation:

N_A(t) = N_A(0) e^{-\lambda t}

where N_A(t) is the number of nuclei $A$ remaining at time $t$, N_A(0) is the initial number of $A$ nuclei, and \lambda is the decay constant.

The decay constant \lambda is related to the half-life T by:

\lambda = \frac{\log(2)}{T}

The number of nuclei $B$ at time $t$ is the number of $A$ nuclei that have decayed. Therefore:

N_B(t) = N_A(0) - N_A(t)

N_B(t) = N_A(0) - N_A(0) e^{-\lambda t} = N_A(0)(1 - e^{-\lambda t})

We are given that the ratio \frac{N_B(t)}{N_A(t)} = 0.3. Plugging in the expressions, we get:

\frac{N_A(0)(1 - e^{-\lambda t})}{N_A(0)e^{-\lambda t}} = 0.3

\frac{1 - e^{-\lambda t}}{e^{-\lambda t}} = 0.3

1 - e^{-\lambda t} = 0.3 e^{-\lambda t}

e^{-\lambda t} = \frac{1}{1.3}

Taking the natural logarithm on both sides, we have:

-\lambda t = \log\left(\frac{1}{1.3}\right) = -\log(1.3)

\lambda t = \log(1.3)

Substitute the expression for \lambda:

\frac{\log(2)}{T} t = \log(1.3)

Solving for t:

t = T \frac{\log (1.3)}{\log (2)}

Therefore, the correct answer is:

t = T \frac{\log \, (1.3)}{\log \, 2}