Question

A and B are the two radioactive elements. The mixture of these elements show a total activity of 1200 disintegrations/minute. The half life of A is 1 day and that of B is 2 days. What will be the total activity after 4 days? Given: The initial number of atoms in A and B are equal.
A. 200dis/min
B.250dis/min
C. 500dis /min
D. 150 dis/min

Answer 1

Hint: If two radioactive substances are mixed, their total activity would be the sum of individual activities. The activity, which is defined as the rate at which disintegrations are happening is given as $R = \dfrac{{dN}}{{dt}} = \lambda N$.

Complete step-by-step answer:
Let us consider a mixture of two radioactive elements $A$ and $B$ each with $N$ atoms initially. Let their disintegration constants be ${\lambda _A}$ and ${\lambda _B}$ and the corresponding half-lives be ${T_A}$ and ${T_B}$.
We know that the activity of a sample is the number of disintegrations happening per second.
So we can say activity $R = \dfrac{{dN}}{{dt}} = \lambda N$
This means the activity due to $A$ would be ${\lambda _A}N$ and that of $B$ would be ${\lambda _B}N$
We can say the total activity, which given is :
${\lambda _A}N + {\lambda _B}N = 1200$ eqn(1)
Let’s recall from the relation for half-life that disintegration constant $\lambda$ and Half life $T$are related as:
$\lambda = \dfrac{{ln(2)}}{T}$
Since we are given the half life of $A$ and $B$, we can find their $\lambda$ as :
${\lambda _A} = \dfrac{{ln(2)}}{{{T_A}}} = \dfrac{{ln(2)}}{{1\;day}}$
${\lambda _B} = \dfrac{{ln(2)}}{{{T_B}}} = \dfrac{{ln(2)}}{{2\;day}}$

Let’s substitute these values into eqn (1) So as to find N.
$\dfrac{{ln(2)}}{1}N + \dfrac{{ln(2)}}{2}N = 1200$
$\dfrac{{3\ln \left( 2 \right)}}{2}N = 1200$
$N = \dfrac{{1200 \times 2}}{{3ln(2)}}$

Now, since the half life of A is 1day and that of B is 2 days, after four days A would have undergone 4 half lives and B would have undergone 2 half lives.
So the number of atoms of A left after four days would be
$\dfrac{N}{{2 \times 2 \times 2 \times 2}} = \dfrac{N}{{16}}$
Similarly that of B left would be :
$\dfrac{N}{{2 \times 2}} = \dfrac{N}{4}$

So we see that the final activity would be
${R_f} = {\lambda _A}{N_A} + {\lambda _B}{N_B}$=${\lambda _A}\dfrac{N}{{16}} + {\lambda _B}\dfrac{N}{4}$
Let’s substitute the values of ${\lambda _A}$, ${\lambda _B}$ and $N$.
${R_f} = \left( {\dfrac{{ln(2)}}{{1day}}} \right)\dfrac{N}{{16}} + \left( {\dfrac{{ln(2)}}{{2day}}} \right)\dfrac{N}{4}$

${R_f} = N\left( {\left( {\dfrac{{ln(2)}}{{1day}}} \right)\dfrac{1}{{16}} + \left( {\dfrac{{ln(2)}}{{2day}}} \right)\dfrac{1}{4}} \right)$
${R_f} = \dfrac{{1200 \times 2}}{{3ln(2)}}\left( {\left( {\dfrac{{ln(2)}}{{1day}}} \right)\dfrac{1}{{16}} + \left( {\dfrac{{ln(2)}}{{2day}}} \right)\dfrac{1}{4}} \right)$
${R_f} = \dfrac{{1200 \times 2}}{3}\left( {\dfrac{1}{{16}} + \dfrac{1}{8}} \right)$
${R_f} = \dfrac{{1200 \times 2}}{3}\dfrac{3}{{16}}$
${R_f} = \dfrac{{1200}}{8} = 150dis/min$
This is the required answer.

Note: We can simplify the calculations for competitive exams by considering just the ratios.
We know half the lives of A and B are in the ratio 1:2. Since $R = \lambda N = \dfrac{{ln(2)N}}{T}$, R and T are inversely proportional. This means activities of A and B are in the ratio 2:1.
So if 1200 disintegrations happen in a minute, 800 of them are from A and 400 are from B.
After 4 days, A would have undergone 4 half lives and B would have undergone 2.
So their final activities would be $\dfrac{{800}}{{{2^4}}}$ and$\dfrac{{400}}{{{2^2}}}$ respectively.
So the total activity is $50 + 100 = 150$dis/min.