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
    $4860$

Answer

Correct option: A.
$1080$
a
(a) $\lambda = {\lambda _1} + {\lambda _2} \Rightarrow \frac{1}{T} = \frac{1}{{{T_1}}} + \frac{1}{{{T_2}}}$

$\therefore T = \frac{{{T_1}{T_2}}}{{{T_1} + {T_2}}} = \frac{{810 \times 1620}}{{810 + 1620}} = 540\;years$

Hence $\frac{1}{4} \,th$ of material remain after $1080$ years.

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