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When working with light, understanding how to convert between wavelength and frequency is critical. Wavelength, denoted by \( \lambda \), is the distance between successive peaks of a wave. Frequency, \( f \), is how often these peaks pass a particular point per second. These two properties are related by the speed of light.

The relationship is captured in the formula:

• \( \lambda = \frac{c}{f} \)

where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \). To find the frequency given the wavelength, rearrange the formula to:

• \( f = \frac{c}{\lambda} \)

For light of wavelength 620 nm, convert this to meters first (\( 620 \times 10^{-9} \text{ m} \)) to use in our equation. The resulting frequency is approximately \( 4.84 \times 10^{14} \text{ Hz} \). This step is important for energy calculations in physics and chemistry, linking to Planck's equation for energy values.

Planck's equation is fundamental in quantum mechanics and is used to link the energy of a photon to its frequency. The formula is:

• \( E = hf \)

where \( E \) is the energy of the photon in joules, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s} \)), and \( f \) is the frequency determined from the wavelength.

By substituting the frequency \( 4.84 \times 10^{14} \text{ Hz} \) into the equation, the energy comes out to approximately \( 3.21 \times 10^{-19} \text{ J} \). This calculation allows us to determine how much energy a photon carries, which is crucial, for example, in determining the energy gaps in semiconductors.

This energy can then be used further in conversions to practical units like electronvolts.

In physics, energy is often converted into electronvolts (eV) for convenience. One electronvolt is the amount of kinetic energy gained by an electron when it is accelerated through a potential difference of one volt. The conversion factor between joules and electronvolts is:

• \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \)

To convert the energy obtained from Planck's equation, \( 3.21 \times 10^{-19} \text{ J} \), into electronvolts, use the formula:

• \( E_g = \frac{E}{1.602 \times 10^{-19}} \)

This gives an energy gap \( E_g \) of approximately \( 2.00 \text{ eV} \). This step is crucial for interpreting energy levels in a way that's compatible with electronic and solid-state physics experiments. Understanding this conversion helps bridge theoretical calculations with practical applications in electronics.