College

The rate of decay of a radioactive element at a given instant of time is [tex]10^3[/tex] disintegrations per second. If the half-life of this element is 1 second, then the rate of decay after 3 seconds will be:

A) 12 per sec
B) 500 per sec
C) 50 per sec
D) 125 per sec

One-eighth of the initial mass of a certain radioactive isotope remains undecayed after a given period of time. The half-life of the isotope in minutes is:

Answer :

To solve this problem, we'll use the formula that describes radioactive decay. The rate of decay of a radioactive element can be determined by the following exponential decay formula:

[tex]\[ N(t) = N_0 \cdot e^{-\lambda t} \][/tex]

Where:
- [tex]\( N(t) \)[/tex] is the quantity that remains after time [tex]\( t \)[/tex].
- [tex]\( N_0 \)[/tex] is the initial quantity or the initial rate of decay.
- [tex]\( \lambda \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time elapsed.
- [tex]\( e \)[/tex] is the base of the natural logarithm.

Given Data:
- Initial decay rate ([tex]\( N_0 \)[/tex]) = [tex]\(10^3\)[/tex] disintegrations per second
- Half-life ([tex]\( t_{1/2} \)[/tex]) = 1 second
- Time elapsed ([tex]\( t \)[/tex]) = 3 seconds

### Step-by-Step Solution:

1. Determine the Decay Constant [tex]\( \lambda \)[/tex]:

The decay constant [tex]\( \lambda \)[/tex] is related to the half-life by the formula:

[tex]\[ \lambda = \frac{\ln(2)}{t_{1/2}} \][/tex]

Given that the half-life [tex]\( t_{1/2} \)[/tex] is 1 second:

[tex]\[
\lambda = \frac{\ln(2)}{1} \approx 0.693
\][/tex]

2. Calculate the Rate of Decay After 3 Seconds:

We use the decay formula [tex]\( N(t) = N_0 \cdot e^{-\lambda t} \)[/tex]:

- Initial decay rate [tex]\( N_0 = 10^3 \)[/tex] disintegrations per second
- Decay constant [tex]\( \lambda \approx 0.693 \)[/tex]
- Time elapsed [tex]\( t = 3 \)[/tex] seconds

Substitute these values into the decay formula:

[tex]\[
N(3) = 10^3 \cdot e^{-0.693 \cdot 3}
\][/tex]

Using the property of exponents:

[tex]\[
e^{-0.693 \cdot 3} = (e^{-0.693})^3 \approx \left(\frac{1}{2}\right)^3 = \frac{1}{8}
\][/tex]

Therefore:

[tex]\[
N(3) = 10^3 \cdot \frac{1}{8} = 125 \text{ disintegrations per second}
\][/tex]

Based on the detailed calculations, the rate of decay after 3 seconds is [tex]\(\boxed{125 \text{ per second}}\)[/tex].

So the correct answer is [tex]\( \boxed{D \text{ (125 per sec)}} \)[/tex].