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: Number of half lives is inversely linked to the half life:
In a fixed time, a sample with a shorter half life goes through more half lives. So $T\propto 1/n$.

Step 2: Count n:
A: $n_A = 3$. B: $n_B = 7$ (because $2^7 = 128$).

Step 3: Ratio:
$\dfrac{T_B}{T_A} = \dfrac{n_A}{n_B} = \dfrac37$.

Step 4: Check:
$T_A = 3$ h and $T_B = 9/7 = 1.29$ h. B has the shorter half life and decays further, as stated.

Final Answer:
Option (C). \[ \boxed{3:7 \text{ (C)}} \]