Question:
A radioactive sample contains \(10^{-3}\,kg\) each of two nuclear species A and B with half-life 4 days and 8 days respectively. The ratio of the amounts of A and B after 16 days is
A radioactive sample contains \(10^{-3}\,kg\) each of two nuclear species A and B with half-life 4 days and 8 days respectively. The ratio of the amounts of A and B after 16 days is
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Shorter half-life → faster decay → smaller remaining quantity.
Updated On: May 8, 2026
- 1 : 2
- 4 : 1
- 1 : 4
- 2 : 1
- 1 : 1
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The Correct Option is C
Solution and Explanation
Concept:
Radioactive decay:
\[
N = N_0 \left(\frac{1}{2}\right)^{t/T}
\]
Step 1: For substance A.
\[ T_A = 4\,days,\quad t = 16\,days \] \[ N_A = N_0 \left(\frac{1}{2}\right)^{16/4} = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16} \]
Step 2: For substance B.
\[ T_B = 8\,days \] \[ N_B = N_0 \left(\frac{1}{2}\right)^{16/8} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4} \]
Step 3: Take ratio. \[ \frac{N_A}{N_B} = \frac{1/16}{1/4} = \frac{1}{4} \] \[ N_A : N_B = 1 : 4 \] \[ \boxed{1:4} \]
Step 1: For substance A.
\[ T_A = 4\,days,\quad t = 16\,days \] \[ N_A = N_0 \left(\frac{1}{2}\right)^{16/4} = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16} \]
Step 2: For substance B.
\[ T_B = 8\,days \] \[ N_B = N_0 \left(\frac{1}{2}\right)^{16/8} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4} \]
Step 3: Take ratio. \[ \frac{N_A}{N_B} = \frac{1/16}{1/4} = \frac{1}{4} \] \[ N_A : N_B = 1 : 4 \] \[ \boxed{1:4} \]
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