**Question:**

A radioactive isotope has a half-life of *T *years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

**Solution:**

Half-life of the radioactive isotope = *T* years

Original amount of the radioactive isotope = *N*0

(a) After decay, the amount of the radioactive isotope = *N*

It is given that only 3.125% of *N*0 remains after decay. Hence, we can write:

$\frac{N}{N_{0}}=3.125 \%=\frac{3.125}{100}=\frac{1}{32}$

But $\frac{N}{N_{0}}=e^{-\lambda t}$

Where,

λ = Decay constant

*t* = Time

$\therefore-\lambda t=\frac{1}{32}$

$-\lambda t=\ln 1-\ln 32$

$-\lambda t=0-3.4657$

$t=\frac{3.4657}{\lambda}$

Since $\lambda=\frac{0.693}{T}$

$\therefore t=\frac{3.466}{\frac{0.693}{T}} \approx 5 T$ years

Hence, the isotope will take about 5*T* years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = *N*

It is given that only 1% of *N*0 remains after decay. Hence, we can write:

$\frac{N}{N_{0}}=1 \%=\frac{1}{100}$

But $\frac{N}{N_{0}}=\mathrm{e}^{-\lambda t}$

$\therefore \mathrm{e}^{-\lambda t}=\frac{1}{100}$

$-\lambda t=\ln 1-\ln 100$

$-\lambda t=0-4.6052$

$t=\frac{4.6052}{\lambda}$

Since, *λ* = 0.693/*T*

*$*\therefore t=\frac{4.6052}{\frac{0.693}{T}}=6.645 T$ years

Hence, the isotope will take about 6.645*T* years to reduce to 1% of its original value.

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