Question

At some instant two radioactive substances are having amount in ratio 2:1.Their half-lives 12 hr and 16 hr, then after two days what will be the ratio if their quantities?

Answer 1

We will take hours as units for time t and decays per hour (3600 Bq) as unit for decay constant for the solution below. (instead of per second).

Here, when we say an amount of substance, it means the number of Nuclei or atoms.

$\frac{N}{N_0} = e^{-\lambda\ t},\ \ \ \ \ \ T_{\frac{1}{2}} = \frac{ Ln\ 2}{\lambda},\ \ \ \ \ \ \lambda = \frac{Ln\ 2}{ T_{\frac{1}{2}}} \\ \\ \lambda_1 = \frac{Ln\ 2}{12 } /hour,\ \ \ \ \ \lambda_2 = \frac{Ln\ 2}{16 } /hour \\ \\ N1_0 = 2\ N2_0 \\ \\ Ln\ \frac{N_1}{ N1_{0}} = -\frac{Ln\ 2}{12} * t,\ \ \ \ \ \ \ Ln\ \frac{N_2}{ N2_{0}} = -\frac{Ln\ 2}{16} * t\\ \\Subtract\ 2nd\ equation\ from\ 1st\ equation \\ \\Ln\ \frac{N_1}{ N1_{0}} - Ln\ \frac{N_2}{ N2_{0}} = [-\frac{Ln\ 2}{12}+\frac{Ln\ 2}{16}] * t$


$Ln\ [\frac{N_1}{ N1_{0}}* \frac{N2_0}{ N_2}] = -\ (Ln\ 2) * \frac{1}{48} * t\\ \\Ln\ [\frac{N1}{2*N2}] = -(\frac{1}{48}\ Ln\ 2) t\\ \\ -Ln\ 2+Ln\frac{N1}{N2}=-(\ (Ln\ 2) * \frac{1}{48}) * t \\ \\Ln\frac{N1}{N2}= Ln\ 2 * [1-\frac{t}{48}]$

After t = 48 hours, the ratio of the amounts of the two radioactive substances is 

$Ln\frac{N1}{N2}= (Ln\ 2) * [1-\frac{48}{48}] = 0\\ \\ \frac{N1} {N2} = 1$

The amounts of the two substance will become equal. That is both substances will have same number of (atoms or) Nuclei.

Answer 2

Use the formula  a  = A x (1/2)^(t/half-life)

Where A = initial quantity,  a = amount after decay in time t
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Let there be 2n and n initial quantities.

For the first substance, A = 2n, half-life  = 12 hrs,  t = 48 hrs
==> a = (2n)(1/2)^(48/12) = 2xnx(1/2)^4 = 2xnx(2)^(-4) = n x 2^(-3) ---------(1)
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For the second substance, A = n, half-life = 16 hrs, t = 48 hrs
==> a = n x (1/2)^(48/16) = n x (1/2)^3 = n x (2)^(-3) -----------------------------(2)
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From (1) and (2),  ratio of the two quantities will be 1 : 1, i.e., both substances will have equal quantity after 48 hrs.