KCET 2015 Physics Solved Paper

Given, half-life of radioactive substance, T

1

2

=20 minutes
Using ⇒t=−‌

1

λ

‌ln(‌

N

N0

)
We know ‌

1

λ

=‌

T1∕2

ln‌2

t=‌

−ln‌2

T1∕2

‌ln(‌

N

N0

)
If t1 is time for 50% decay and t2 is time for 87.5% decay, we have
=‌

T1∕2

ln‌2

‌ln(‌

N1

N2

)
Substituting the values, we get Therefore, time taken between 50% decay and 87.5% decay will be 40 minutes.