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. $150dis/min$

Answer 1

Answer: (D)

$150dis/min$

Let the initial number of atoms be $N_o$.

In $4days$, $A$ has four half lives.

Atoms of $A$ remaining after four days, $N_A=\frac{N_o}{2^4}=\frac{N_o}{16}$

In $4days$, $B$ has two half lives.

Hence, atoms of $B$ remaining after two days, $N_B=\frac{N_o}{2^2}=\frac{N_o}{4}$

Total activity of an element is given by $A=\left|\frac{dN}{dt}\right|=\lambda N$

Initial activity is $1200={\lambda }_A{N}_o+{\lambda }_B{N}_o.............\left(i\right)$

Half life is given by:

$t_{1/2}=\frac{\ln 2}{\lambda }$

$⟹{\lambda }_A=\frac{\ln 2}{1440}...............\left(ii\right)$ ($1day=60\times 24min$)

${\lambda }_B=\frac{\ln 2}{2\times 1440}.............\left(iii\right)$

Using (i), (ii) and (iii),

$N_o=\frac{1200\times 1440\times 2}{3\ln 2}...............\left(iv\right)$

After $4days$, activity is $A=\frac{{\lambda }_A{N}_o}{16}+\frac{{\lambda }_B{N}_o}{4}..............\left(v\right)$

Substituting (ii), (iii) and (iv) in (v),

$A=\frac{\frac{\ln 2}{1440}\times \frac{1200\times 1440\times 2}{3\ln 2}}{16}+\frac{\frac{\ln 2}{2\times 1440}\times \frac{1200\times 1440\times 2}{3\ln 2}}{4}$

$A=150dis/min$