Question

A radioactive nucleus X with a half-life \(T\), decays into a nucleus Y. At \(t=0\), there is no nucleus Y. After some time '\(t\)', the ratio of the number of Y to that of X is \(0.4\). Then, '\(t\)' is

• A. \(T\log (1.4)\)
• B. \(\frac{T\log (1.4)}{\log (2)}\)
• C. \(\frac{T\log (2)}{2\,\log (1.4)}\)
• D. \(\frac{T}{\log (2)}\)

Hint:

Write the number of X and Y nuclei with the exponential law and set Y/X equal to 0.4.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let the initial number of X nuclei be \(N_0\). After time \(t\), \(N_X=N_0e^{-\lambda t}\) and \(N_Y=N_0-N_X=N_0\left(1-e^{-\lambda t}\right)\).

Step 2: Use the given ratio
\(\dfrac{N_Y}{N_X}=\dfrac{1-e^{-\lambda t}}{e^{-\lambda t}}=e^{\lambda t}-1=0.4\).

Step 3: Solve for time
\(e^{\lambda t}=1.4\Rightarrow \lambda t=\ln 1.4\). With \(\lambda=\dfrac{\ln 2}{T}\), \(t=\dfrac{T\ln 1.4}{\ln 2}\).

Step 4: Compare the options
This is option (B), \(\dfrac{T\log(1.4)}{\log(2)}\).

Final Answer:
The time is T log(1.4) divided by log 2. \[ \boxed{\text{(B) }\dfrac{T\log (1.4)}{\log (2)}} \]