A radioactive material decays by simulataneous emission of two particles with half - lives 1620 and 810 years. The time after which one - fourth of the material remains, is

1080 years

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2430 years

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3240 years

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4860 years

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Solution

The correct option is A
1080 years

$λ = λ_{1} + λ_{2} = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{t_{\frac{1}{2} (1)}} + \frac{0.693}{t_{\frac{1}{2} (2)}} t_{\frac{1}{2}} = \frac{t_{\frac{1}{2} (1)} \times t_{\frac{1}{2} (2)}}{t_{\frac{1}{2} (1)} + t_{\frac{1}{2} (2)}} = \frac{1620 \times 810}{1620 + 810} = 540 y e a r s F r a c t i o n l e f t = \frac{1}{4} = (\frac{1}{2})^{2}$ No. of half-lives = 2
Hence, time required $= 2 \times 540 = 1080$ year

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A radioactive material decays by simulataneous emission of two particles with half - lives 1620 and 810 years. The time after which one - fourth of the material remains, is

The correct option is A 1080 years λ=λ1+λ2=0.693t12=0.693t12(1)+0.693t12(2)t12=t12(1)×t12(2)t12(1)+t12(2)=1620×8101620+810=540 yearsFraction left=14=(12)2 No. of half-lives = 2 Hence, time required =2×540=1080 year