# For the Lyman series, n_f = 1. How can we calculate the E_"photon" for the bandhead of the Lyman series (the transition n = oo -> n = 1 for emission) in joules and in eV?

## The same energy is needed for the transition $n = 1 \to n = \infty$, which is the ionization potential for a hydrogen atom.

Nov 25, 2017

$2.17 \cdot {10}^{- 18}$ $\text{J}$

$\text{13.6 eV}$

#### Explanation:

Your tool of choice here will be the Rydberg equation

$\frac{1}{l a m \mathrm{da}} = R \cdot \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

Here

• $l a m \mathrm{da}$ si the wavelength of the emittted photon
• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$
• ${n}_{f}$ is the final energy level of the transition
• ${n}_{i}$ is the initial energy level of the transition

In your case, you have the

${n}_{i} = \infty \to {n}_{f} = 1$

transition, which is part of the Lyman series. The first thing to notice here is that when ${n}_{i} = \infty$

$\frac{1}{n} _ {i}^{2} \to 0$

which implies that the Rydberg equation can be simplified to this form

$\frac{1}{l} a m \mathrm{da} = R \cdot \left(\frac{1}{1} ^ 2 - 0\right)$

$\frac{1}{l a m \mathrm{da}} = R$

You can thus say that the wavelength of the emitted photon will be equal to

$l a m \mathrm{da} = \frac{1}{R}$

lamda = 1/(1.097 * 10^7color(white)(.)"m"^(-1)

$l a m \mathrm{da} = 9.158 \cdot {10}^{- 8} \textcolor{w h i t e}{.} \text{m}$

Now, to find the energy of the photon emitted during this transition, you can use the Planck - Einstein relation

$E = \frac{h \cdot c}{l} a m \mathrm{da}$

Here

• $E$ is the energy of the photon
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} \textcolor{w h i t e}{.} \text{J s}$
• $c$ is the speed of light in a vacuum, usually given as $3 \cdot {10}^{8} \textcolor{w h i t e}{.} {\text{m s}}^{- 1}$

Plug in your value to find

E = (6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(9.158 * 10^(-8)color(red)(cancel(color(black)("m"))))

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{E = 2.17 \cdot {10}^{- 18} \textcolor{w h i t e}{.} \text{J}}}}$

I'll leave the answer rounded to three sig figs.

To convert this to electronvolts, use the fact that

$\text{1 eV" = 1.6 * 10^(-19)color(white)(.)"J}$

You will end up with

$2.17 \cdot {10}^{- 18} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) * "1 eV"/(1.6 * 10^(-19)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)("13.6 eV}}}}$

This basically means that you need $\text{13.6 eV}$ to ionize a hydrogen atom. In other words, the first energy level of a hydrogen atom, i.e. its ground state, is at $- \text{13.6 eV}$.

If an incoming photon has an energy that is at least $\text{13.6 eV}$, then the electron will move to an energy level that is high enough to be considered outside the influence of the nucleus, and thus outside the atom $\to$ the hydrogen atom is ionized to a hydrogen cation, ${\text{H}}^{+}$.