For the Lyman series, $n_{f} = 1$ $n_f = 1$ . How can we calculate the $E_{photon}$ $E_"photon"$ for the bandhead of the Lyman series (the transition $n = \infty \to n = 1$ $n = oo -> n = 1$ for emission) in joules and in eV?

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For the Lyman series, $n_{f} = 1$ $n_f = 1$ . How can we calculate the $E_{photon}$ $E_"photon"$ for the bandhead of the Lyman series (the transition $n = \infty \to n = 1$ $n = oo -> n = 1$ for emission) in joules and in eV?

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

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Answer 1 · 2017-11-25T00:34:46

$2.17 \cdot 10^{- 18}$ $2.17 * 10^(-18)$ $J$ $"J"$

$13.6 eV$ $"13.6 eV"$

Explanation:

Your tool of choice here will be the Rydberg equation

$\frac{1}{λ} = R \cdot (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ $1/(lamda) = R * (1/n_f^2 - 1/n_i^2)$

Here

$λ$ $lamda$ si the wavelength of the emittted photon

$R$ $R$ is the Rydberg constant, equal to $1.097 \cdot 10^{7}$ $1.097 * 10^(7)$ $m^{- 1}$ $"m"^(-1)$

$n_{f}$ $n_f$ is the final energy level of the transition

$n_{i}$ $n_i$ is the initial energy level of the transition

In your case, you have the

$n_{i} = \infty \to n_{f} = 1$ $n_i = oo -> n_f = 1$

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

$\frac{1}{n_{i}^{2}} \to 0$ $1/n_i^2 -> 0$

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

$\frac{1}{λ} = R \cdot (\frac{1}{1^{2}} - 0)$ $1/lamda = R * (1/1^2 - 0)$

$\frac{1}{λ} = R$ $1/(lamda) = R$

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

$λ = \frac{1}{R}$ $lamda = 1/R$

$λ = \frac{1}{1.097 \cdot 10^{7} . m^{- 1}}$ $lamda = 1/(1.097 * 10^7color(white)(.)"m"^(-1)$

$λ = 9.158 \cdot 10^{- 8} . m$ $lamda = 9.158 * 10^(-8)color(white)(.)"m"$

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

$E = \frac{h \cdot c}{λ}$ $E = (h * c)/lamda$

Here

$E$ $E$ is the energy of the photon

$h$ $h$ is Planck's constant, equal to $6.626 \cdot 10^{- 34} . J s$ $6.626 * 10^(-34)color(white)(.)"J s"$

$c$ $c$ is the speed of light in a vacuum, usually given as $3 \cdot 10^{8} . {m s}^{- 1}$ $3 * 10^8color(white)(.)"m s"^(-1)$

Plug in your value to find

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

$color(darkgreen)(ul(color(black)(E = 2.17 * 10^(-18)color(white)(.)"J")))$

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

To convert this to electronvolts, use the fact that

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

You will end up with

$2.17 * 10^(-18) color(red)(cancel(color(black)("J"))) * "1 eV"/(1.6 * 10^(-19)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)("13.6 eV")))$

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

![https://chemistry.stackexchange.com/questions/44625/what-is-the-series-limit-in-a-spectral-line]()

If an incoming photon has an energy that is at least $"13.6 eV"$ , 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 $->$ the hydrogen atom is ionized to a hydrogen cation, $"H"^(+)$ .

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For the Lyman series, $n_{f} = 1$ $n_f = 1$ . How can we calculate the $E_{photon}$ $E_"photon"$ for the bandhead of the Lyman series (the transition $n = \infty \to n = 1$ $n = oo -> n = 1$ for emission) in joules and in eV?

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

1 Answer

Explanation:

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

The same energy is needed for the transition n = 1 -> n = oon=1→n=∞n = 1 -> n = oo, which is the ionization potential for a hydrogen atom.

1 Answer

Explanation:

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Impact of this question

For the Lyman series, $n_{f} = 1$ $n_f = 1$ . How can we calculate the $E_{photon}$ $E_"photon"$ for the bandhead of the Lyman series (the transition $n = \infty \to n = 1$ $n = oo -> n = 1$ for emission) in joules and in eV?

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