Question

A radioactive substance is found to register 5000 counts per second on a Geiger counter. Twenty-four hours later it registers 1250 counts per second. What is its half-life?

Answer 1

The half-life of the radioactive substance = 12 hours.

Step-by-step explanation

Given that:

R₀ = 5000 counts/sec

R₁ = 1250 counts/sec

t = 24 hr

What to find:

The half-life of the radioactive substance.

Step-by-step solution:

The half-life of life of the radioactive substance can be calculated using the formula below:

`\frac{0.693t}{T_{1\text{/}2}}=ln((R_0)/(R_1))`

Putting the values of the parameters into the formula, we have:

`\begin{gathered} \frac{0.693*24}{T_{1\text{/}2}}=ln((5000)/(1250)) \\ \\ \frac{16.632}{T_{1\text{/}2}}=ln(4) \\ \\ \frac{16.632}{T_{1\text{/}2}}=1.3863 \\ \\ T_{1\text{/}2}=(16.632)/(1.3863) \\ \\ T_{1\text{/}2}=11.997\text{ }hr \\ \\ T_{1\text{/}2}\approx12\text{ }hr \end{gathered}`

Thus, the half-life of the radioactive substance is 12 hours.