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

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: $$\Delta E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{589.6 \times 10^{-9}} \approx 2.1 \, \text{eV}$$