Question

The half-life of a radioactive substance is 5 hours. A sample is tested and found to contain 0.48 of the substance. How much of the substance was present in the sample 20 hours before the sample was tested

Answer 1

Answer:

$X_f=0.03$

Explanation:

The half-life of a substance is defined by:

$\displaystyle \frac{X_f}{X_i}=\left(\frac{1}{2}\right)^{t/\tau}$

Xf: The final number of substance left

Xi: The initial number of substance

t: Time elapsed

$\tau$ (or commonly used $t_{1/2}$): Half-life

From here, we know:

$\tau=5$

$X_i=0.48$

$t=20$

$Xf=?$

With all the information given, we can now solve for Xf:

$\displaystyle \frac{X_f}{0.48}=\left(\frac{1}{2}\right)^{20/5}$

$\displaystyle X_f=0.48 \cdot \left(\frac{1}{2} \right)^4$

$X_f=0.03$

After 20 hours, 0.03 of the substance will be present.