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

To find the energy difference between the two energy levels in sodium responsible for emitting a spectral line at 589.6 nm, we use the relationship between energy, frequency, and wavelength given by the formula:

Energy (E) = h × c / λ

Where:

• h = Planck's constant = 6.626 × 10^-34 J·s
• c = Speed of light in a vacuum = 3.00 × 10⁸ m/s
• λ = Wavelength = 589.6 nm = 589.6 × 10^-9 m

Substitute these values into the formula to find the energy:

E = (6.626 × 10^-34 J·s) × (3.00 × 10⁸ m/s) / (589.6 × 10^-9 m)

E ≈ 3.37 × 10^-19 J

To convert this energy from joules to electronvolts (eV), use the conversion factor:

• 1 eV = 1.602 × 10^-19 J

Thus,

E ≈ 3.37 × 10^-19 J / 1.602 × 10^-19 J/eV

E ≈ 2.1 eV

Therefore, the difference in energy between the two energy levels is 2.1 eV, which matches the correct answer.

Answer 2

The energy difference \( \Delta E \) between two energy levels is related to the wavelength \( \lambda \) of the emitted light by the equation: \[ \Delta E = \frac{hc}{\lambda} \] Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) is Planck's constant,
- \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light,
- \( \lambda = 589.6 \, \text{nm} = 589.6 \times 10^{-9} \, \text{m} \). Substituting the values: \[ \Delta E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{589.6 \times 10^{-9}} \approx 2.1 \, \text{eV} \]