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?

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

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?

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

 

[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.