Are these expressions equal?

120 Views Asked by At

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?

2

There are 2 best solutions below

1
On BEST ANSWER

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!

0
On

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...