pGRE question on natural line width The lifetime for the $2p \rightarrow 1s$ transition in hydrogen is $1.6 \times 10^{-9}$ s. The natural line width for the radiation emitted during the transition is approximately...
Their solution:

The natural line width, $\Gamma$, is \begin{align}\Gamma &= \frac{\hbar}{\tau}
\\&= \frac{\hbar}{1.6 \times 10^{-9}}{\rm Hz} \\&= h \cdot \frac{1}{2 \pi \cdot 1.6 \cdot 10^{-9}} {\rm Hz} \\&\approx h \cdot 10^8 {\rm Hz}\end{align}
Therefore, the answer (C), 100MHz, is correct.

I don't understand why the $h$ isn't multiplied. $100 {\rm\,MHz} = 10^8\,{\rm Hz}$, but leaving the $h$ there is like multiplying by $10^{-34}$, and none of the answers have an $h$ in them.
What am I missing?
 A: The line width in the formula is calculated in Joules and in your case $\Gamma = 6.6\cdot10^{-26}\rm J$. It is an awkward order of magnitude (nobody knows the prefix for $10^{-27}$) so spectroscopists just prefer to use implicit energy units. In your case it is $\rm MHz$, but there are more strange and still commonly used units like a wavenumber $(\rm cm^{-1})$ which is calculated by dividing the frequency by the speed of light expressed in $\rm cm/s$! In your case, $$\Gamma = h\cdot c \cdot 0.0033\,\rm cm^{-1}$$ 
It is done for convenience of working with the numbers in the reasonable range (0.001 to 1000). As a side effect you always have to remember to multiply the number by a proper constant when you put it in calculations.
A: So really this comes from one of the pseudo-Heisenberg uncertainty relations where:
$$\Delta E\times \Delta t=\hbar$$
And the subsequent uncertainty is in units of energy. If you then compare that to the equation for energy of a photon:
$$E=\hbar \times\nu$$
It's pretty easy to see that you can substitute in $E$ for $\Delta E$. The frequency you solve for will be a $\Delta \nu$ which corresponds to the uncertainty in energy and your $\hbar$s cancel. So you wind up with that answer in frequency because it's a question about spectroscopy and thus you want something related to photon energy. Also it's an answer you can get in your head fairly quickly. Your natural linewidth equation seems to be missing some things.
