A radioactive material decays by simultaneous emission of two particles with respective half lives 1

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13.Nuclei

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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

A

$1080$

B

$2430$

C

$3240$

D

$4860$

(IIT-1995) (AIIMS-2008)

Solution

(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.

Standard 12

Physics

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