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-life of the radioisotope?

Respuesta :

11.54 minutes

Explanation:

The decay rate equation is given by

[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]

where [tex]\lambda[/tex] is the half-life. We can rewrite this as

[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]

Taking the natural logarithm of both sides, we get

[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]

Solving for [tex]\lambda[/tex],

[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]

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

[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]