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Question

# A radioactive material by simultaneous emission of two particles with respective half-lives 1620 and 810 year. The time, in years, after which one-fourth of the martial remains is:

A
1080
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B
2430
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C
3240
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D
4860
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Solution

## The correct option is A 1080From Rutherford-Soddy law, the number of atoms left after n half-lives is given by N=N0(12)nwhere N0 is original number of atoms.The number of half-life n=timeofdecayefectivehalf−lifeRelation between effective disintegration constant(λ) and half- life(T) is λ=ln2TTherefore, λ1+λ2=ln2T1+ln2T2Effective half- life 1T=1T1+1T2=11620+1810 1T=1+21620 T=540yrTherefore, n=t540 N=N0(12)t540 NN0=(12)2=(12)t540 ⇒t540=2 ⇒t=2×540=1080yr

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