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?




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.


 5

This issue is explicitly addressed in BIPM: *[The International System of Units (SI)](http://www.bipm.org/en/publications/si-brochure/)* 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, $2.51\times10^{11}\ \mathrm{ms}^{-1}=2.51\times10^{11}\ (\mathrm{ms})^{-1}$.


 2