Question

In a time of \(10\) minutes, the activity of a radioactive sample becomes \[ \frac{1}{\sqrt5} \] times its initial activity. After \(10\) more minutes, if its activity becomes \(K\) times the initial activity, the value of \(K\) is

• A. \(0.2\)
• B. \(5\)
• C. \(0.5\)
• D. \(2\)

Hint:

Radioactive activity follows \[ \boxed{ A=A_0e^{-\lambda t}. } \] Equal successive time intervals multiply the activity by the same decay factor.

Answer 1

The Correct Option is A

Step 1: Use the radioactive decay law. Activity is given by \[ A=A_0e^{-\lambda t}. \] After \(10\) minutes, \[ \frac{A}{A_0} = \frac1{\sqrt5}. \]

Step 2: Find the activity after another \(10\) minutes. After a total of \(20\) minutes, \[ \frac{A}{A_0} = \left(\frac1{\sqrt5}\right)^2 = \frac15 = 0.2. \] Hence, \[ K=0.2. \] Therefore, \[ \boxed{(A)} \] is the correct answer.