Question

Sodium atoms emit a spectral line with a wavelength in the yellow, 589.6 nm. What is the difference in energy between the two energy levels involved in the emission of this spectral line?

• A. 2.6 eV
• B. 2.9 eV
• C. 2.1 eV
• D. None

Hint:

The energy of a photon can be calculated using \( \Delta E = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength of the emitted light.

Answer 1

The Correct Option is C

Step 1: Find the frequency of the light.
The wavelength is \( \lambda = 589.6 \, \text{nm} = 589.6 \times 10^{-9} \, \text{m} \). Using \( c = \nu \lambda \), the frequency is \( \nu = \dfrac{c}{\lambda} = \dfrac{3 \times 10^8}{589.6 \times 10^{-9}} \approx 5.09 \times 10^{14} \, \text{Hz} \).

Step 2: Apply the Planck relation.
The energy of a photon is \( E = h\nu \), where \( h = 6.626 \times 10^{-34} \, \text{J s} \). So \( E = 6.626 \times 10^{-34} \times 5.09 \times 10^{14} \approx 3.37 \times 10^{-19} \, \text{J} \).

Step 3: Convert to electronvolts.
Since \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \), \( E = \dfrac{3.37 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.1 \, \text{eV} \).
\[ \boxed{E \approx 2.1 \, \text{eV}} \]

Answer 2

Instead of working in joules and converting afterward, we can use Planck's constant expressed directly in electronvolt-seconds, \( h = 4.136 \times 10^{-15} \, \text{eV s} \), and see which of the given energies is consistent with the quoted wavelength.

1. 2.6 eV: Using \( E = \dfrac{hc}{\lambda} \) rearranged as \( \lambda = \dfrac{hc}{E} \), an energy of 2.6 eV would need \( \lambda = \dfrac{4.136 \times 10^{-15} \times 3 \times 10^8}{2.6} \approx 477 \times 10^{-9} \, \text{m} \), far shorter than 589.6 nm.
2. 2.9 eV: The same substitution gives \( \lambda \approx 428 \times 10^{-9} \, \text{m} \), shorter still, so this energy also does not fit the given wavelength.
3. 2.1 eV: Substituting \( E = 2.1 \) eV gives \( \lambda = \dfrac{4.136 \times 10^{-15} \times 3 \times 10^8}{2.1} \approx 590.9 \times 10^{-9} \, \text{m} \), which is very close to the stated 589.6 nm.
4. None: Because 2.1 eV reproduces the correct wavelength almost exactly, this option is ruled out.

Only the 2.1 eV value reconstructs the given wavelength when run back through \( E = \dfrac{hc}{\lambda} \) using \( h \) in eV s.

So the correct answer is 2.1 eV.