Question

The count rate from a radioactive sample falls from 4.0 × 106 per second to 1.0 × 106 per second in 20 hours. What will be the count rate 100 hours after the beginning?

Answer 1

$3.9 \times 10^3$ disintegrations per second

Explanation:

We know that,

$A = A_0e^{-\lambda t}$

$\lambda = \frac {ln2}{half life}$

where  

• A= Activity of the substance = $1 \times 10^6 s^-^1$

• $A_0$ = Initial activity = $4 \times 10^6 s^-^1$

• t = time = 20h

$\lambda$ = decay constant

So according to the problem,

$4 \times 10^6 \times e^{-\lambda t} = 1 \times 10^6 s^-^1$

$e^{-20\lambda} = 0.25$

Now we have to evaluate activity after 100 hours

A = $4 \times 10^6 \times e^{-100\lambda}$

A = $4 \times 10^6 \times (e^{-20\lambda})^5$

A = $4 \times 10^6 \times (0.25)^5$

A = $3.9 \times 10^3$ disintegrations per second

Hence, the activity after 100 hours is $3.9 \times 10^3$ disintegrations per second