# Does the prefix of an inverted unit also get inverted?

What is the energy of radiation that has a frequency of $$\pu{2.51 \times 10^11 ms-1}$$?

(a) $$\pu{1.66 \times 10^-19 J}$$ (supposedly correct)
(b) $$\pu{1.66 \times 10^-22 J}$$
(c) $$\pu{7.92 \times 10^-37 J}$$
(d) $$\pu{1.66 \times 10^-25 J}$$

My argument was that: since $$\pu{Hz} = \pu{s-1}$$, then $$\pu{ms-1} = \pu{mHz}$$. So, I divided $$2.51 \times 10^{11}$$ by $$1000$$ and solved the question normally (using $$E = h\nu$$). But my professor said that since $$\pu{m}$$ is a prefix, it should follow whatever is in front of it, so $$\pu{ms-1}$$ will become $$1/\pu{ms}$$, then you multiply it by 1000 to change it to Hz. Is that the correct way to do it?

• I applaud the literal interpretation, but yes the prefix goes with the unit therefore ms^-1 is equivalent to (ms)^-1. – A.K. Apr 11 '19 at 14:13
• The prefix is part of the unit, in that milliseconds is its own unit. However, in this situation, it seems silly to write $\mathrm{ms}^{-1}$ when you could instead write $10^{3}\ \mathrm{s}^{-1}$. – Zhe Apr 11 '19 at 17:58
• Yes it does, but to remember it in this form means to sweep a misunderstanding under the rug, which invites troubles in the future. Think of it this way. There are no prefixes. There are just thousands, millions, and other numbers, which you supposedly know how to handle when they turn $(\dots)^{-1}$. – Ivan Neretin Apr 12 '19 at 4:59

You should read $$\pu{ms-1}$$ as $$(\pu{ms})^{-1}$$ and not as $$\mathrm{m}(\pu{s-1})$$. Your teacher is correct.

[...] since $$\pu{Hz} = \pu{s-1}$$ then $$\pu{ms-1} = \pu{mHz}$$

This is incorrect.

If something happens once per hour, it happens less often than when it happens once per minute. This is because an hour is longer than a minute. In a similar manner, if something happens once per second, it happens less often than when it happens once per millisecond. This is because a second is longer than a millisecond.

So $$\pu{ms-1} = \pu{kHz}$$, i.e. the frequency of something happening every millisecond is larger than the frequency of something happening every second.

This issue is explicitly addressed in BIPM: The International System of Units (SI) as follows.

The grouping formed by a prefix symbol attached to a unit symbol constitutes a new inseparable unit symbol (forming a multiple or submultiple of the unit concerned) that can be raised to a positive or negative power and that can be combined with other unit symbols to form compound unit symbols.

Examples: $$2.3\ \mathrm{cm}^3=2.3\ (\mathrm{cm})^3=2.3\ (10^{-2}\ \mathrm m)^3=2.3\times10^{-6}\ \mathrm m^3$$ $$1\ \mathrm{cm}^{-1}=1\ (\mathrm{cm})^{-1}=1\ (10^{-2}\ \mathrm m)^{-1}=10^2\ \mathrm m^{-1}=100\ \mathrm m^{-1}$$ $$1\ \mathrm{V/cm}=(1\ \mathrm V)/(10^{-2}\ \mathrm m)=10^2\ \mathrm{V/m}=100\ \mathrm{V/m}$$ $$5000\ \mathrm{µs}^{-1}=5000\ (\mathrm{µs})^{-1}=5000\ (10^{-6}\ \mathrm s)^{-1}=5\times10^9\ \mathrm s^{-1}$$

Therefore, $$2.51\times10^{11}\ \mathrm{ms}^{-1}=2.51\times10^{11}\ (\mathrm{ms})^{-1}$$.