Question

One mole of radium has an activity of \( \frac{1}{6.3 \times 10^{37}} \) kilo curie. Its decay constant is:

• A. \( \frac{1}{6} \times 10^{-10} \, {s}^{-1} \)
• B. \( 10^{-10} \, {s}^{-1} \)
• C. \( 10^{-11} \, {s}^{-1} \)
• D. \( 10^{-8} \, {s}^{-1} \)

Hint:

Remember that Avogadro's number represents the number of particles in one mole of any substance, and is crucial in calculations involving mole-based quantities in nuclear physics.

Answer 1

The Correct Option is A

Step 1: The activity \( A \) of a radioactive substance is given by the formula: \[ A = \lambda N \] where \( \lambda \) is the decay constant and \( N \) is the number of nuclei. For one mole of radium, \( N \) is Avogadro's number (\( N = 6 \times 10^{23} \)). 
Step 2: The activity is given as \( A = \frac{1}{6.3 \times 10^{37}} \) kilo curie. We convert this into becquerels (1 kilo curie = \( 3.7 \times 10^{10} \) Bq), so: \[ A = \frac{1}{6.3 \times 10^{37}} \times 3.7 \times 10^{10} = \frac{1}{6.3 \times 10^{34}} \, {Bq} \] Step 3: Now, using the formula for the decay constant: \[ \lambda = \frac{A}{N} = \frac{\frac{1}{6.3 \times 10^{34}}}{6 \times 10^{23}} \approx \frac{1}{6} \times 10^{-10} \, {s}^{-1} \]