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What is the wavelength (in nanometers) of the photon emitted when an electron in a hydrogen atom transitions from the ni=5 energy level to the nf=2 energy level?

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Question

What is the wavelength (in nanometers) of the photon emitted when an electron in a hydrogen atom transitions from the \(n_i​=5\) energy level to the \(n_f​=2\) energy level?

Interpretation

This question involves the Bohr model of the hydrogen atom, which states that an electron can occupy only specific, quantized energy levels. When an electron moves from a higher energy level to a lower energy level, it loses energy. This lost energy is emitted as a photon.

The energy of each level in the hydrogen atom is given by the Bohr energy equation. The energy change for a transition is\( ΔE = R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2}) \)

where

  • \(RH​=2.18×10−18 J\) is the Rydberg energy constant,
  • \(ni​\) is the initial energy level,
  • \(nf​\) is the final energy level.

A negative value of \(ΔE\) indicates that energy is released (emission), while a positive value indicates that energy is absorbed (absorption).

Once the energy of the photon is known, its wavelength can be calculated using

\( E = \frac{hc}{λ} \)

or

\(λ = \frac{hc}{E} \)

Concept Application

In this problem, the electron moves from\(n_i=5\rightarrow n_f=2\)

Since the electron moves to a lower energy level, this is an emission process.

Although the calculated energy change will be negative, the wavelength of a photon is always positive because wavelength is a physical distance. Therefore, while calculating the wavelength, we use only the magnitude of the energy.

Solution

The energy change is

\( ΔE = R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2}) \)

Substituting the given values,

\( ΔE = (2.18\times10^{-18}(\frac{1}{5^2}-\frac{1}{2^2}) \)

\( = (2.18\times10^{-18}(\frac{1}{25}-\frac{1}{4}) \)

\( = (2.18\times10^{-18})(0.04 – 0.25) \)

\( = (2.18\times10^{-18})(-0.21) \)

\( ΔE = -4.58\times10^{-19}J \)

The negative sign confirms that the photon is emitted.

To calculate the wavelength, use the magnitude of the energy:

\(λ = \frac{hc}{|ΔE|} \)

Substituting the known constants,

λ=(3×108ms−1)(6.63×10−34Js)4.58×10−19Jλ=\frac{(3\times10^{8}\,\mathrm{m\,s^{-1}})(6.63\times10^{-34}\,\mathrm{J\,s})}{4.58\times10^{-19}\,\mathrm{J}}

\( λ = 4.34\times10^{-7}m \)

Finally, convert meters to nanometers:

λ=4.34×10−7m×1nm10−9m=434nm\lambda=4.34\times10^{-7}\,\text{m}\times\frac{1\,\text{nm}}{10^{-9}\,\text{m}}=434\,\text{nm}

Final Answer

\(\ λ=434nm\)

​

Insight

A simple way to remember electronic transitions is:

  • Electron moves down (\(ni​>nf​\)) → Energy is emitted → \(ΔE<0\)
  • Electron moves up (\(ni​<nf​\)) → Energy is absorbed → \(ΔE>0\)

Also remember that energy and wavelength are inversely related:

\( E = \frac{hc}{λ} \)

This means that a larger energy difference produces a shorter wavelength, while a smaller energy difference produces a longer wavelength. This relationship is fundamental to understanding atomic spectra and is frequently tested in chemistry examinations.

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