# What is the difference in energy between the lowest energy state of the hydrogen atom, and the next higher energy level (i.e. the difference in energy between the 2 lowest energy states)?

May 18, 2016

As a preface, since you have asked for the "difference in energy", I will give an answer that has no sign. That is, I will give $| \Delta E |$, a magnitude.

Since you are looking at the Bohr model of the hydrogen atom, recall that there exists something called the Rydberg equation (or formula):

$\setminus m a t h b f \left(\frac{1}{\lambda} = {R}_{H} \left(\frac{1}{{n}_{f}^{2}} - \frac{1}{{n}_{i}^{2}}\right)\right)$

where:

• $\lambda$ is the wavelength in $\text{m}$,corresponding to the difference in energy between quantum levels ${n}_{i}$ and ${n}_{f}$.
• $n$ is the principal quantum number as usual.
• ${R}_{H}$ is the Rydberg constant, ${\text{10973731.6 m}}^{- 1}$.

(Normally there's a sign in the equation, but all we need to do is ensure that $\lambda > 0$.)

The lowest energy state is for hydrogen's electron in $n = 1$, and the next-highest state is $n = 2$. So, let ${n}_{f} = 2$ and ${n}_{i} = 1$.

To solve for the difference in energy, we will also need another equation.

If you recall, the difference in energy is related to the frequency (which is related to the wavelength) by Planck's constant.

$\setminus m a t h b f \left(\Delta E = h \nu = \frac{h c}{\lambda}\right)$

where:

• $\Delta E$ is the difference in energy you're looking for.
• $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant.
• $\nu$ is the frequency corresponding to the difference in energy.
• $c = 2.998 \times {10}^{8} \text{m/s}$ is the speed of light.

So, combining these two equations, we would proceed as follows:

$\lambda = \frac{h c}{\Delta E} ,$

$\frac{1}{\lambda} = \frac{1}{\frac{h c}{\Delta E}} = {R}_{H} \left(\frac{1}{{n}_{f}^{2}} - \frac{1}{{n}_{i}^{2}}\right)$

$\frac{\Delta E}{h c} = {R}_{H} \left(\frac{1}{{n}_{f}^{2}} - \frac{1}{{n}_{i}^{2}}\right)$

$\setminus m a t h b f \left(\Delta E = h c {R}_{H} \left(\frac{1}{{n}_{f}^{2}} - \frac{1}{{n}_{i}^{2}}\right)\right)$

The difference in energy, $| \Delta E |$, for a single hydrogen atom, is:

$\textcolor{b l u e}{| \Delta E |} = h c {R}_{H} | \frac{1}{{n}_{f}^{2}} - \frac{1}{{n}_{i}^{2}} |$

= (6.626xx10^(-34) "J"cdotcancel"s")(2.998xx10^8 cancel"m""/"cancel"s")("10973731.6" cancel("m"^(-1)))
$\times | \frac{1}{{2}^{2}} - \frac{1}{{1}^{2}} |$

$= \textcolor{b l u e}{1.635 \times {10}^{- 18} \text{J}}$

Or, if you wanted this in $\text{eV}$, which is how it is often referenced, use the conversion factor of $1.602 \times {10}^{- 19} \text{J/eV}$ to get:

color(blue)(|DeltaE|) = 1.635xx10^(-18) "J" xx "1 eV"/(1.602xx10^(-19) "J")

$\approx$ $\textcolor{b l u e}{\text{10.2 eV}}$

And indeed, the lowest energy level of hydrogen atom ($n = 1$) is $- \text{13.6 eV}$, and that at $n = 2$ is about $- \text{3.4 eV}$.