MCQ
A radioactive material decays by simultaneous emission of two particles with respective half lives $1620$ and $810$ years. The time (in years) after which one- fourth of the material remains is
- ✓$1080$
- B$2430$
- C$3240$
- D$4860$
$\therefore T = \frac{{{T_1}{T_2}}}{{{T_1} + {T_2}}} = \frac{{810 \times 1620}}{{810 + 1620}} = 540\;years$
Hence $\frac{1}{4} \,th$ of material remain after $1080$ years.
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