Ncert Exercises & Solutions

13.7 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?

Hint: Activity at any time t is given by, \(\frac{\mathrm{R}}{\mathrm{R}_{0}}=\mathrm{e}^{-\lambda t}\).
(a)

Step 1: Find the time after which the activity reduces to 3.125% of its original value.
The half-life of the radioactive isotope =T years and the original amount of the radioactive isotope = N₀
After decay, the activity of the radioactive isotope=R

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 \(\frac{\mathrm{R}}{\mathrm{R}_{0}}=\mathrm{e}^{-\lambda t}\)

,
where

λ

= Decay constant and t = Time
So,
\(\mathrm{e}^{-\lambda t}=\frac{1}{32}\)

- λt = \ln 1 - \ln 32

- λt = 0 - 3.4657

t = \frac{3.4657}{λ}

Since λ = \frac{0.693}{T}

∴ t = \frac{3.466}{\frac{0.693}{T}} \approx 5 T

Years

Hence, the isotope will take about 5T years to reduce to 1% of its original value.
(b)
Step 2: Find the time after which the activity reduces to 1% of its original value.

After decay, the activity of the radioactive isotope=R

It is given that only 1% of

N_{0}

remains after decay. Hence, we can write:

\frac{R}{R_{0}} = 1 % = \frac{1}{100}

But \frac{N}{N_{0}} = e^{- λt}

∴ e^{- λt} = \frac{1}{100}

- λt = \ln 1 - \ln 100

- λt = 0 - 4.6052

t = \frac{4.6052}{λ}

Since, λ = 0.693 / T

∴ t = \frac{4.6052}{\frac{0.693}{T}} = 6.645 T years

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

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