Question

Radioactive isotope has a half-life of t years after how much time its activity reduces to 6.25% of its original activity

Answer 1

answer : 4t years

given, half life of a radioactive isotope = t years

we know, half life = ln2/λ, where λ is radioactive constant.

so, t = ln2/λ ⇒λ = ln2/t

now using formula, t' = 1/λ ln{a/a-x}

here a - x = 6.25 % of a = 6.25a/100 = 0.0625a

so, t' = t/ln2 × ln(1/0.0625) = t/ln2 × ln(16)

= t × 4ln2/ln2

= 4t

hence, time taken to reduce to 6.25% of its original activity is 4t years.

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Answer 2

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A radioactive isotope has a half life of T years how long will it take the activity to reduce to 3.125% of its original value

Thus, the isotope will take about 5T years to reduce to 3.125% of its original value. Hence, the isotope will take about 6.645Tyears to reduce to 1% of its original value.