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zimovet [89]
3 years ago
9

A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes la

ter is 1,000 counts per minute. What is the half-life of the radioisotope?
Physics
1 answer:
zmey [24]3 years ago
4 0

11.54 minutes

Explanation:

The decay rate equation is given by

N = N_0e^{-\frac{t}{\lambda}}

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

\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}

Taking the natural logarithm of both sides, we get

\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)

Solving for \lambda,

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

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

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

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3. The load across a 12-V battery consists of a series combination
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