-3
$\begingroup$

In my book it says

Convert $1.213\cdot10^{-11}$ metre into Angstrom.

The answer is $12.13\cdot10^{-2}$ Angstrom.

How is it possible? I thought of multiplying the metre with $10^{-10}$, but I didn't get the correct answer. The same applies to the conversion of centimetres to Angstrom.

$\endgroup$
3
$\begingroup$

The ångström (symbol: Å) is a widely used non-SI unit of length. Its value in SI units is:

$$1\ \mathring{\mathrm{A}}=0.1\ \mathrm{nm}=100\ \mathrm{pm}=10^{-10}\ \mathrm m$$

Thus, the required conversion factor is given by

$$\frac{1\ \mathring{\mathrm{A}}}{10^{-10}\ \mathrm m}=1$$

or

$$\frac{10^{-10}\ \mathrm m}{1\ \mathring{\mathrm{A}}}=1$$

This conversion factor can be used to express the given length of $1.213\cdot10^{-11}\ \mathrm m$ in terms of the unit ångström:

$$1.213\cdot10^{-11}\ \mathrm m=1.213\cdot10^{-11}\ \mathrm m\cdot\frac{1\ \mathring{\mathrm{A}}}{10^{-10}\ \mathrm m}=0.1213\ \mathring{\mathrm{A}}$$

Note that $0.1213\ \mathring{\mathrm{A}}=12.13\cdot10^{-2}\ \mathring{\mathrm{A}}$, which is the answer given in your book.

Also note that you do not simply divide the given length by $10^{-10}$ since the units are part of the conversion factor. You actually divide the value by $10^{-10}\ \mathrm m/\mathring{\mathrm{A}}$.

| improve this answer | |
$\endgroup$
1
$\begingroup$

You're correct in saying you need to do is multiply by $10^{-10}$. The answer you gave is equivalent to $1.123\cdot10^{-1}$ (just divide by $10$), which is exactly $(1.123\cdot10^{-11}) \cdot (10^{-10})$. If you're using a calculator, make sure you wrap each term in brackets so that the order of operations is correct.

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.