Question

Two samples A and B contain equal amount of radioactive substances. If \((\frac{1}{8})^{th}\) of sample A and \((\frac{1}{128})^{th}\) of sample B, remain after 9 hours, then the ratio of half life period of B to that of A is

• A. \(9:7\)
• B. \(7:3\)
• C. \(3:7\)
• D. \(3:1\)

Hint:

Fraction left is (1/2)^(t/T). 1/8 means 3 half lives and 1/128 means 7 half lives in the same 9 hours.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
After \(n\) half lives, the fraction of a radioactive sample that remains is \(\left(\dfrac12\right)^n\), where \(n = \dfrac{t}{T_{1/2}}\).

Step 2: Key Formula or Approach:
Find the number of half lives for each sample in 9 hours, then divide 9 hours by that number.

Step 3: Detailed Explanation:
Sample A: \(\dfrac18 = \left(\dfrac12\right)^3\), so \(n_A = 3\) and
\[ T_A = \frac{9}{3} = 3 \text{ h} \]
Sample B: \(\dfrac{1}{128} = \left(\dfrac12\right)^7\), so \(n_B = 7\) and
\[ T_B = \frac97 \text{ h} \]
Ratio:
\[ \frac{T_B}{T_A} = \frac{9/7}{3} = \frac{9}{21} = \frac37 \]
Option (B) 7:3 is the inverse ratio of the half lives (it is the ratio of numbers of half lives, \(n_B : n_A\)). Option (A) 9:7 is \(T_A:\)... using the 9 hours without dividing by 3.

Final Answer:
\(T_B : T_A = 3:7\), option (C). \[ \boxed{3:7 \text{ (C)}} \]