The half-life of a radioactive substance is $30$ minutes. The time (in minutes) taken between $40$ % decay and $85$ % decay of the same radioactive substance is

Question

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Answer 1

$N_{0}$ =Nuclei at time $t = 0$
$N_{1}$ =Remaining nuclei after $40$ % decay $= (1 - 0.4) N_{0} = 0.6 N_{0}$
$N_{2}$ =Remaining nuclei after $85$ % decay
$= (1 - 0.85) N_{0} = 0.15 N_{0} ∴ \frac{N _{2}}{N _{1}} = \frac{0 . 1 5 N _{0}}{0 . 6 N _{0}} = \frac{1}{4} = (\frac{1}{2})^{2}$
Hence, two half life is required beweeen $40$ % decay and $85$ % decay of radio active substance.
Time taken $= 2 π \frac{1}{2} = 2 \times 30 = 60 min$

Answer 2

N_{0}

=Nuclei at time

t = 0

N_{1}

=Remaining nuclei after

40

% decay

= (1 - 0.4) N_{0} = 0.6 N_{0}

N_{2}

=Remaining nuclei after

85

% decay

= (1 - 0.85) N_{0} = 0.15 N_{0} ∴ \frac{N _{2}}{N _{1}} = \frac{0 . 1 5 N _{0}}{0 . 6 N _{0}} = \frac{1}{4} = (\frac{1}{2})^{2}

Hence, two half life is required beweeen

40

% decay and

85

% decay of radio active substance.

Time taken

= 2 π \frac{1}{2} = 2 \times 30 = 60 min