Question

A radioactive isotope is being produced at a constant rate \(X\). Half-life of the radioactive substance is \(Y\). After some time the number of radioactive nuclei becomes constant. The value of this constant is:

• A. \(\dfrac{XY}{\ln(2)}\)
• B. \(XY\)
• C. \((XY)\ln(2)\)
• D. \(\dfrac{X}{Y}\)

Hint:

In radioactive equilibrium: \[ \text{Constant number of nuclei} = \frac{\text{Production rate}}{\text{Decay constant}} \] Always convert half-life to decay constant using \(\lambda = \dfrac{\ln(2)}{T_{1/2}}\).

Answer 1

The Correct Option is A

Step 1: Understand the physical situation. The radioactive nuclei are:
being produced at a constant rate \(X\),
decaying simultaneously. After a long time, a steady state is reached where: \[ \text{Rate of production} = \text{Rate of decay} \]
Step 2: Write the decay law. Let \(N\) be the constant number of radioactive nuclei. Rate of decay: \[ \text{Decay rate} = \lambda N \] where \(\lambda\) is the decay constant.
Step 3: Express decay constant in terms of half-life. \[ \lambda = \frac{\ln(2)}{T_{1/2}} = \frac{\ln(2)}{Y} \]
Step 4: Apply steady-state condition. \[ X = \lambda N \] \[ N = \frac{X}{\lambda} \] Substitute \(\lambda = \dfrac{\ln(2)}{Y}\): \[ N = \frac{X}{\ln(2)/Y} = \frac{XY}{\ln(2)} \]
Hence, the constant number of radioactive nuclei is \[ \boxed{\dfrac{XY}{\ln(2)}} \]