Question

A radioactive material decreases by simultaneous emissions of two particles with half life 1620 and 810 years. The time after which (1/4)th of the material remained is:

• A. 1080 years
• B. 2000years
• C. 1500years
• D. 1200years

Answer 1

The correct option is A 1080 years

$A\underrightarrow{k_1}B$

$A\underrightarrow{k_2}C$

$-\frac{dA}{dt}=(k_1+k_2)\left[A\right]=k\left[A\right]$

$t_{1/2}$ overall = $\frac{(t_{1/2}{)}_1\times (t_{1/2}{)}_2}{(t_{1/2}{)}_1+(t_{1/2}{)}_2}=\frac{1620\times 810}{1620+810}=540$ years

Overall $k=\frac{In2}{540}$

Now, $N=N_0{e}^{-kt}$

given $N=\frac{N_0}{4}$
$e^{kt}=4$

$kt=2In2$

$t=\frac{2In2}{In2}\times t_{1/2}=2\times 540=1080$ years