Photon Energy & Wavelength

PHYAtoms — Bohr Model & Spectra

When an electron in a hydrogen atom jumps between Bohr levels, it emits or absorbs a single photon whose energy equals the energy gap, $E_{photon} = E_{n_2} - E_{n_1}$ , and whose wavelength follows $\lambda = hc/E$ . The Rydberg formula packages every line of the hydrogen spectrum into one expression $1/\lambda = R(1/n_1^2 - 1/n_2^2)$ , sorting the lines into the Lyman, Balmer and Paschen series by their lower level.

ISC examines this constantly because it links Bohr energies, photon energy and spectral wavelength, and the shortcut $hc = 1240\,\text{eV}\cdot\text{nm}$ converts an energy in eV straight into a wavelength in nm.

Photon energy and wavelength

E_{photon} = E_{n_2} - E_{n_1} = h\nu = \dfrac{hc}{\lambda}, \qquad \lambda\,(\text{nm}) = \dfrac{1240}{E\,(\text{eV})}

E_{photon}

is the energy gap between the two levels,

h

Planck's constant,

\nu

frequency,

\lambda

wavelength,

c

the speed of light. For the

n=3 \to n=2

jump,

E = -1.51-(-3.4)=1.89\,\text{eV}

, giving

\lambda = 1240/1.89 \approx 656\,\text{nm}

(the

\text{H}\alpha

line).

Rydberg formula for hydrogen spectral lines

\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right), \qquad R = 1.097\times 10^{7}\ \text{m}^{-1}

\lambda

is the emitted wavelength,

R

the Rydberg constant,

n_1

the lower (final) level and

n_2 > n_1

the higher (initial) level. The series is set by

n_1

: Lyman

n_1=1

, Balmer

n_1=2

, Paschen

n_1=3

Series limit (shortest wavelength of a series)

\dfrac{1}{\lambda_{min}} = \dfrac{R}{n_1^{2}} \quad (n_2 \to \infty)

\lambda_{min}

is the series limit, reached when the upper level

n_2 \to \infty

1/n_2^2 \to 0

. For the Lyman series (

n_1=1

\lambda_{min} = 1/R \approx 91.2\,\text{nm}

— the shortest-wavelength (most energetic) line of that series.

Longest wavelength of a series (first member)

\dfrac{1}{\lambda_{max}} = R\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{(n_1+1)^{2}}\right)

\lambda_{max}

is the longest-wavelength (lowest-energy) line, the transition from the level just above the lower one,

n_2 = n_1+1

. For Paschen (

n_1=3

), the

4\to 3

line is the longest-wavelength member.

Emission means a jump down ( $n_2 \to n_1$ , photon released); absorption means a jump up. The photon energy equals the magnitude of the energy gap either way.
Smaller energy gap means longer wavelength: within a series the first member ( $n_2 = n_1+1$ ) is the longest wavelength, and the series limit ( $n_2 \to \infty$ ) is the shortest.
Series are named by their final level: Lyman ( $n_1=1$ , UV), Balmer ( $n_1=2$ , visible), Paschen ( $n_1=3$ , IR). Only the Balmer series falls in the visible band.
The $\text{H}\alpha$ line is the Balmer first member, $n=3\to 2$ , at about $656\,\text{nm}$ (red); it is the most-quoted single spectral line in the chapter.
Keep $R$ in $\text{m}^{-1}$ and the answer comes out in metres; convert to nm or Å at the end ( $1\,\text{nm}=10^{-9}\,\text{m}$ , $1\,\text{Å}=10^{-10}\,\text{m}$ ).
The shortcut $hc = 1240\,\text{eV}\cdot\text{nm}$ is exact enough for ISC: divide it by the photon energy in eV to get $\lambda$ directly in nm.
The Rydberg and Bohr-energy routes agree because $R$ is built from the same constants as the $13.6\,\text{eV}$ ground-state energy; either method gives the same wavelength.

Where the marks go

Swapping $n_1$ and $n_2$ in the Rydberg formula, which makes $1/\lambda$ negative. Always put the smaller (lower) level as $n_1$ so the bracket stays positive.
Confusing 'shortest wavelength' with 'longest': the series limit ( $n_2\to\infty$ ) is the shortest wavelength, the first member ( $n_1+1\to n_1$ ) is the longest.
Mixing the energy-gap shortcut with the wrong unit — $hc = 1240$ works only for $E$ in eV and $\lambda$ in nm; in SI use $\lambda = hc/E$ with $hc = 1.986\times 10^{-25}\,\text{J}\cdot\text{m}$ .
Putting the Balmer series in the UV — only the Balmer series is visible; Lyman is UV and Paschen is IR.

How the board asks it

Numericalthe rydberg formula and $\lambda = hc/E$
The electron in a hydrogen atom jumps from the $n=3$ level to the $n=2$ level. Taking $R = 1.097\times 10^{7}\,\text{m}^{-1}$ , calculate the wavelength of the emitted photon and state the series and spectral region to which it belongs.
Numericalthe energy-gap shortcut $hc = 1240\,\text{eV}\cdot\text{nm}$
The ground-state energy of hydrogen is $-13.6\,\text{eV}$ . Calculate the energy of the photon emitted when the electron de-excites from $n=2$ to $n=1$ , and hence find its wavelength in $\text{nm}$ .
Numericalseries limit and first member of a series
For the Lyman series of hydrogen, calculate (i) the longest wavelength (first member) and (ii) the series limit (shortest wavelength), taking $R = 1.097\times 10^{7}\,\text{m}^{-1}$ .
Derive / provebohr energies linking to the rydberg expression
Using Bohr's expression for the energy of the $n^{\text{th}}$ level of hydrogen, derive an expression for the wavelength of the radiation emitted when the electron jumps from level $n_2$ to level $n_1$ , and identify the Rydberg constant.
Give reasonsseries are named by their final level; only balmer is visible
Account for the fact that the lines of the Balmer series of hydrogen lie in the visible region whereas those of the Lyman series lie in the ultraviolet.
Give reasonssmaller energy gap means longer wavelength
The $\text{H}\alpha$ line ( $n=3\to 2$ ) of the Balmer series has a longer wavelength than the $\text{H}\beta$ line ( $n=4\to 2$ ). Give reasons.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.