Question

$A$ and $B$ are two substances undergoing radioactive decay in a container The half life of $A$ is $15 min$ and that of $B$ is $5\, min$ If the initial concentration of $B$ is $4$ times that of $A$ and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? ______ min

Hint:

Whensolvingproblemsinvolvingradioactivedecay, it’simportanttousethefor mulaforexponentialdecayandbalancetheexponentstofindtherequiredtime

Answer 1

Correct Answer: 15

Calculation of Time for Equal Concentrations of A and B: 

The decay of a substance follows the formula:

N(t) = N₀ × (1/2)^t/t_1/2

Where:

• N(t): Concentration at time t
• N₀: Initial concentration
• t_1/2: Half-life of the substance

Let the initial concentration of A be N_A, and the initial concentration of B be N_B = 4N_A.

1. For substance A:
   N_A(t) = N_A × (1/2)^t/15
2. For substance B:
   N_B(t) = 4N_A × (1/2)^t/5
3. Set the concentrations equal:
   N_A × (1/2)^t/15 = 4N_A × (1/2)^t/5
4. Cancel N_A from both sides:
   (1/2)^t/15 = 4 × (1/2)^t/5
5. Rewrite 4 as 2²:
   (1/2)^t/15 = (1/2)^{t/5 - 2}
6. Equating the exponents:
   t/15 = t/5 - 2
7. Solve for t:
   t/15 - t/5 = -2
   Multiply through by 15 to eliminate the fractions:
   t - 3t = -30
   -2t = -30
   t = 15 minutes

Conclusion: It will take 15 minutes for the concentrations of A and B to become the same.

Answer 2

The correct answer is 15.
$[A{]}_t=[A{]}_0{e}^{-kt}$
For $A$ : Let $[A{]}_t$ be $y$ and $[A{]}_0$ be $x;k=\frac{\ln 2}{t_{1/2}}=$
$\frac{\ln 2}{15\min }$
$y=x{e}^{-kt}$
$=x{e}^{-\left(\frac{\ln 2}{15}\right)t}$
For B: $[B{]}_t=[B{]}_0{e}^{-kt}$
Let $[B{]}_t=y;[B{]}_0=4x;k=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{5\min }$
$y=4x{e}^{-\left(\frac{\ln 2}{5}\right)t}$
$\Rightarrow x{e}^{-\left(\frac{\ln 2}{15}\right)t}=4x{e}^{-\left(\frac{\ln 2}{5}\right)t}$
$e^{t\left(\frac{\ln 2}{5}\frac{\ln 2}{15}\right)}=4$
$t\times \left[\frac{\ln 2}{5}-\frac{\ln 2}{15}\right]=\ln 4$
$t\times \ln 2\left[\frac{1}{5}-\frac{1}{15}\right]=2\ln 2$
$t=15\min$