A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-…

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asked Jan 8, 2022 140k views

1 Answer

Braian Silva · Answer 1 · 2022-01-15T02:36:26+0000

11.54 minutes

Step-by-step explanation:

The decay rate equation is given by

$N = N_0e^{-(t)/(\lambda)}$

where
$\lambda$ is the half-life. We can rewrite this as

$(N)/(N_0) = e^{-(t)/(\lambda)}$

Taking the natural logarithm of both sides, we get

$\ln \left((N)/(N_0)\right) = -\left((t)/(\lambda)\right)$

Solving for
$\lambda$ ,

$\lambda = -(t)/(\ln \left((N)/(N_0)\right))$

$\:\:\:\:= -\frac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}$

$\:\:\:\:=11.54\:\text{minutes}$