I'm reading a physics booklet on radioactivity and at one part during the calculations it says:
$$A(0) = 6.0 \cdot 10^{23} \times \frac {\ln 2}{(8 \times 24 \times 3600)} = N(21) = 6.0 \cdot 10^{23} \cdot {\left(\dfrac{1}{2}\right)}^{21/8}$$
But when I try this on my calculator it doesn't work. Coincidentally, it says that these types of calculations can go wrong easily on a calculator, but is this even correct?
No, it can't be right. Ignoring the common factor of $6\times 10^{23}$, the left-hand side $$ \frac{\ln 2}{\text{some integer}}$$ is transcendental, whereas the right-hand side $$ (\text{some rational})^{\text{another rational}} $$ is algebraic. So they are not mathematically identical.
Actually, they are not even approximately equal -- they differ by about 5 orders of magnitude!
I think the expressions should be like this:
$$A(0) = 6.0 \cdot 10^{23} \times \frac {\ln 2}{(8 \times 24 \times 3600)} => N(21) = 6.0 \cdot 10^{23} \cdot {\left(\dfrac{1}{2}\right)}^{21/8}$$
so first expression is just showing that second expression is true. There is no meaning of equal here
Curious as to how this mistake came about. Textbook authors don't just produce gibberish, right? $\ $ Right?
$6.02\times10^{23}$ is just Avogadro's Number, the count of atoms in a mole.
The number 8 is multiplied by 24 and 3600, implying that 8 is a time in days. It's divided into $\ln2$, which is what one does with a half-life in certain calculations
Then the last two terms make perfect sense: the number of atoms after 21 days is the original number times $\frac{1}{2}$ raised to the power of the number of half-lives The first two terms make sense also, if one assumes that $A(0)$ is the activity in decays/second at time $=0$
The middle equals sign? Got me...
