CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 5. Electrons in Atoms

He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second
energy level (n= 2). This is called the Balmer series. Transitions ending in the ground state (n= 1) are called the
Lyman series. The energies released by these transitions are larger, resulting in spectral lines that show up in the
more energetic ultraviolet region of the spectrum. The transitions referred to as the Paschen series and the Brackett
series release smaller amounts of energy, so these spectral lines appear in the less energetic infrared region.

Watch a simulation of the Bohr model of the hydrogen atom at http://www.dlt.ncssm.edu/core/Chapter8-Atomic_St
r_Part2/chapter8-Animations/ElectronOrbits.html.

Sample Problem 5.3: Spectral Lines of the Hydrogen Atom

In the hydrogen atom, the change in energy (∆E) for then= 4 ton= 2 electron transition is equal to 4.09× 10 −^19 J.
Calculate the wavelength (in nm) of the spectral line that results from this electron transition and identify its color.

Step 1: List the known quantities and plan the problem.

Known

• ∆E= 4.09× 10 −^19 J


• Planck’s constant (h) = 6.626× 10 −^34 J•s


• speed of light (c) = 3.00× 108 m/s


• conversion factor 1 m = 10^9 nm

Unknown

• frequency (ν)


• wavelength (λ)

Apply the equation∆E=hνto solve for the frequency of the emitted light. Then, use the equationc=λνto solve
for the wavelength in m. Finally, convert to nm.

Step 2: Calculate.

ν=

∆E

h

=

4. 09 × 10 −^19 J

6. 626 × 10 −^34 J·s

= 6. 17 × 1014 Hz

λ=

c

ν

=

3. 00 × 108 m/s

6. 17 × 1014 Hz

= 4. 86 × 10 −^7 m

4. 86 × 10 −^7 m×

(

109 nm

1 m

)

=486 nm

Step 3: Think about your result.

The 486 nm spectral line corresponds to a blue-green color. Compared to this transition, then= 3 ton= 2 transition
results in a smaller release of energy. This would correspond to the longer wavelength red spectral line, which has a
lower frequency. Then= 5 ton= 2 andn= 6 ton= 2 transitions presumably correspond to the shorter wavelength
lines.

Practice Problems

4. The energy change (∆E) for then= 2 ton= 1 transition of the Lyman series is 1.64× 10 −^18 J. Calculate the
   wavelength (in nm) of the resulting spectral line.


5. The visible red spectral line that results from then= 3 ton= 2 transition of the hydrogen atom has a wavelength
   of 656 nm. Calculate the energy change that produces this spectral line.