1. 2T

  2. T

  3. 3/2T

  4. 3T

Answers (1)

Radioactive decay, N(t) = N_{0} e^{- \frac{0.693t}{t_{1/2}}}

Where N(t) is the amount of sample which remains after time t.

Given that 75% of the sample has decayed, hence the remaining will be 25% of initial, N(t) =0.25 N_{0}

Also t_{1/2} =T

0.25 N_{0} =N_{0}e^{- \frac{0.693t}{T}}

\Rightarrow 0.25 =e^{- \frac{0.693t}{T}}

Take loge both sides and solve

t = 2T

Option (1) is correct

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