Atoms decay via alpha emission have a half-life of 150 min

[IKonquer]

$$2.3 \cdot 10^{10}$$ atoms decay via alpha emission have a half-life of 150 min.

How many alpha particles are emitted between t=30 min and t=160 min?

$$\begin{flalign*} 150 &= \frac{\ln 2}{\lambda}\\ \lambda &= 0.046 \\ \\ K &= K_{0}e^{(-\lambda)(t)}\\ &= (2.3 \cdot 10^{10})e^{(-0.046)(30)}\\ K_{30} &= 5.79 \cdot 10^{9} \\ \\ K &= K_{0}e^{(-\lambda)(t)}\\ &= (2.3 \cdot 10^{10})e^{(-0.046)(160)}\\ K_{160} &= 1.46 \cdot 10^{7} \\ \\ K_{30} - K_{160} = 5.77 \cdot 10^{9} \text{ emitted} \end{flalign*}$$

Did I do this correctly? And is there a way to use calculus to do this?

 

[SammyS]

λ = ln(2)/150 ≈ 0.004621

You're off by a factor of 10.

K₃₀ represents the number of atoms which remain undecayed after 30 minutes. The number you have for K₃₀ is way under half of the initial sample, in a time that's 1/5 of the half-life.

 

[IKonquer]

Nice catch! Does it look right now? Also is there a way of using integral calculus to solve this?

$$\begin{flalign*} 150 &= \frac{\ln 2}{\lambda}\\ \lambda &= 0.0046 \\ \\ K &= K_{0}e^{(-\lambda)(t)}\\ &= (2.3 \cdot 10^{10})e^{(-0.0046)(30)}\\ K_{30} &= 2.00 \cdot 10^{10} \\ \\ K &= K_{0}e^{(-\lambda)(t)}\\ &= (2.3 \cdot 10^{10})e^{(-0.0046)(160)}\\ K_{160} &= 1.10 \cdot 10^{10} \\ \\ K_{30} - K_{160} = 8.98 \cdot 10^{9} \text{ emitted} \end{flalign*}$$

 

[ideasrule]

Yes, that seems better. I don't know why you want to use integral calculus, since the formula for radioactive decay was derived using integral calculus in the first place.

 

[rwisz]

Yes, your calculations are correct. Calculus can also be used to solve this problem. The rate of decay is given by the differential equation:

$\frac{dN}{dt} = -\lambda N$

where N is the number of atoms at time t and $\lambda$ is the decay constant. The solution to this equation is:

$N(t) = N_{0}e^{-\lambda t}$

where $N_{0}$ is the initial number of atoms.

To find the number of atoms that decay between t=30 min and t=160 min, we can use the definite integral:

$\int_{30}^{160} \frac{dN}{dt} dt = \int_{30}^{160} -\lambda N dt$

Using the solution for N(t), we get:

$\int_{30}^{160} -\lambda N_{0}e^{-\lambda t} dt$

Evaluating the integral, we get:

$N(160) - N(30) = N_{0}e^{-\lambda 160} - N_{0}e^{-\lambda 30}$

Substituting in the given values, we get:

$N(160) - N(30) = (2.3 \cdot 10^{10})e^{-0.046 \cdot 160} - (2.3 \cdot 10^{10})e^{-0.046 \cdot 30}$

$= 1.46 \cdot 10^{7} - 5.79 \cdot 10^{9}$

$= 5.77 \cdot 10^{9}$

So, we get the same answer using calculus as well.