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Question

# The mean lives of radioactive substances are 1620 are 405 years for α−emission and β−emission respectively. Find out the time during which three fourth of a sample will decay if it is decaying both by α−emission and β−emission simultaneously. Choose the approximate answer. Given ln4=1.386.

A
320 Years
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B
449 Years
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C
640 Years
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D
729 Years
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Solution

## The correct option is B 449 YearsWhen a substance decays α and β emission simultaneously, the effective disintegration constant λ is given by λ=λα+λβ Where λα= disintegration constant for α− emission only λβ= disintegration constant for β− emission only Mean life is given by, Tm=1λ, λ=λα+λβ ⇒λ=1Tα+1Tβ ⇒λ=11620+1405=1324 By law of radioactivity, ⇒t=1λln(N0N) Given that, (34)th of the sample is decayed, then the active sample present N=N04 ⇒t=ln4λ=324(1.386)=449 years Hence, option (B) is correct Why this Question? Mean or Average Life (Ta): It is the average of age of all active nuclei, i.e. Ta=sum of times of existence of all nuclei in a sampleinitial number of active nuclei in that sample Ta=1λ

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