Elements like uranium and plutonium are most often studied with half-life in mind.
Therefore, you can calculate the half-life for a particular element and know for certain how quickly it will break down no matter what.
Technically, there are 2 types of carbon: carbon-14, which decays, and carbon-12, which stays constant.
In other words, as x{\displaystyle x} increases, f(x){\displaystyle f(x)} decreases and approaches zero. This is exactly the type of relationship we want to describe half-life. In this case, we want a=12,{\displaystyle a={\frac {1}{2}},} so that we have the relationship f(x+1)=12f(x). {\displaystyle f(x+1)={\frac {1}{2}}f(x). }
f(t)=(12)t{\displaystyle f(t)=\left({\frac {1}{2}}\right)^{t}} Simply replacing the variable doesn’t tell us everything, though. We still have to account for the actual half-life, which is, for our purposes, a constant. We could then add the half-life t1/2{\displaystyle t_{1/2}} into the exponent, but we need to be careful about how we do this. Another property of exponential functions in physics is that the exponent must be dimensionless. Since we know that the amount of substance depends on time, we must then divide by the half-life, which is measured in units of time as well, to obtain a dimensionless quantity. Doing so also implies that t1/2{\displaystyle t_{1/2}} and t{\displaystyle t} be measured in the same units as well. As such, we obtain the function below. f(t)=(12)tt1/2{\displaystyle f(t)=\left({\frac {1}{2}}\right)^{\frac {t}{t_{1/2}}}}
N(t)=N0(12)tt1/2{\displaystyle N(t)=N_{0}\left({\frac {1}{2}}\right)^{\frac {t}{t_{1/2}}}}
Divide both sides by the initial amount N0. {\displaystyle N_{0}. } N(t)N0=(12)tt1/2{\displaystyle {\frac {N(t)}{N_{0}}}=\left({\frac {1}{2}}\right)^{\frac {t}{t_{1/2}}}} Take the logarithm, base 12,{\displaystyle {\frac {1}{2}},} of both sides. This brings down the exponent. log1/2(N(t)N0)=tt1/2{\displaystyle \log {1/2}\left({\frac {N(t)}{N{0}}}\right)={\frac {t}{t_{1/2}}}} Multiply both sides by t1/2{\displaystyle t_{1/2}} and divide both sides by the entire left side to solve for half-life. Since there are logarithms in the final expression, you’ll probably need a calculator to solve half-life problems. t1/2=tlog1/2(N(t)N0){\displaystyle t_{1/2}={\frac {t}{\log {1/2}\left({\frac {N(t)}{N{0}}}\right)}}}
On half-life graphs, the x-axis will usually show the timeline, while the y-axis usually shows the rate of decay.
For example, if the starting point is 1,640, divide 1,640 / 2 to get 820. If you are working with a semi log plot, meaning the count rate is not evenly spaced, you’ll have to take the logarithm of any number from the vertical axis. [11] X Research source
If you know the half-life but you don’t know the initial quantity, you can input the half-life, the quantity that remains, and the time that has passed. As long as you know 3 of the 4 values, you’ll be able to use a half-life calculator.
If you don’t know the half-life but you do know the decay constant and the mean lifetime, you can input those instead. Just like the initial equation, you only need to know 2 of the 3 values to get the third one.
This is a helpful visual, and it can be useful if you don’t want to do all of the equation work.
Solution: we know the initial amount N0=300 g,{\displaystyle N_{0}=300{\rm {\ g}},} final amount N=112 g,{\displaystyle N=112{\rm {\ g}},} and elapsed time t=180 s. {\displaystyle t=180{\rm {\ s}}. } Recall the half-life formula t1/2=tlog1/2(N(t)N0). {\displaystyle t_{1/2}={\frac {t}{\log {1/2}\left({\frac {N(t)}{N{0}}}\right)}}. } Half-life is already isolated, so simply substitute the appropriate variables and evaluate. t1/2=180 slog1/2(112 g300 g)≈127 s{\displaystyle {\begin{aligned}t_{1/2}&={\frac {180{\rm {\ s}}}{\log _{1/2}\left({\frac {112{\rm {\ g}}}{300{\rm {\ g}}}}\right)}}\&\approx 127{\rm {\ s}}\end{aligned}}} Check to see if the solution makes sense. Since 112 g is less than half of 300 g, at least one half-life must have elapsed. Our answer checks out.
Solution: We know the initial amount N0=20 kg,{\displaystyle N_{0}=20{\rm {\ kg}},} final amount N=0. 1 kg,{\displaystyle N=0. 1{\rm {\ kg}},} and the half-life of uranium-232 t1/2=70 years. {\displaystyle t_{1/2}=70{\rm {\ years}}. } Rewrite the half-life formula to solve for time. t=(t1/2)log1/2(N(t)N0){\displaystyle t=(t_{1/2})\log {1/2}\left({\frac {N(t)}{N{0}}}\right)} Substitute and evaluate. t=(70 years)log1/2(0. 1 kg20 kg)≈535 years{\displaystyle {\begin{aligned}t&=(70{\rm {\ years}})\log _{1/2}\left({\frac {0. 1{\rm {\ kg}}}{20{\rm {\ kg}}}}\right)\&\approx 535{\rm {\ years}}\end{aligned}}} Remember to check your solution intuitively to see if it makes sense.
Solution: (1/2)3=0. 125{\displaystyle (1/2)^{3}=0. 125} (the amount remaining after 3 half-lives) 10. 0gx0. 125=1. 25g{\displaystyle 10. 0gx0. 125=1. 25g} remain 10g−1. 25g=8. 75g{\displaystyle 10g-1. 25g=8. 75g} have decayed For this particular equation, the actual length of the half-life did not play a role.
Solution: 17/32=0. 53125{\displaystyle 17/32=0. 53125} (this is the decimal amount that remains) (1/2)n=0. 53125{\displaystyle (1/2)n=0. 53125} nlog0. 5=log0. 53125{\displaystyle nlog0. 5=log0. 53125} n=0. 91254{\displaystyle n=0. 91254} (this is how many half-lives have elapsed) 60min/0. 91254=65. 75min{\displaystyle 60min/0. 91254=65. 75min} n=66min{\displaystyle n=66min} (to 2 sig figs)
