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Planck's equation is crucial for understanding the energy of photons, the fundamental particles of light. According to quantum theory, energy is quantized and can be calculated using the equation (1) Planck's equation is given by:
\( E = hf \)
where \( E \) represents the energy of a photon, \( h \) is Planck's constant with a value of \( 6.626 \times 10^{-34} \text{Js} \), and \( f \) is the frequency of the light wave.
Understanding this equation allows us to link the concept of energy with that of the frequency of electromagnetic waves. The higher the frequency, the greater the energy of the photon. This relationship is fundamental to the photoelectric effect, where electrons are emitted from a material upon the absorption of photons with sufficient energy. By knowing the frequency of the light, Planck's equation can tell us how much energy each photon carries, which in turn is essential in calculating how many photons are necessary to produce a certain amount of energy.

The calculation of photon energy is directly derived from Planck's equation. It is important to do this accurately in photoelectric effect problems because it determines whether the photons have enough energy to eject electrons from a particular material. To determine the energy of a single photon, use the equation:
\( E = hf \), where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is the frequency of the photon.
To calculate the energy of a photon given its frequency, simply multiply the frequency by Planck's constant. For instance, in the original exercise, after determining the frequency of the light, the solution involves the calculation of the energy of a photon:
\( E = (6.626 \times 10^{-34} \mathrm{Js}) \times (6 \times 10^{14} \mathrm{Hz}) \).
Understanding how to calculate photon energy is also helpful in various applications beyond the photoelectric effect, such as in determining the energies required for electronic transitions in atoms, and in the field of spectroscopy, where photon energies are matched to the spectral lines of elements.

While solving photoelectric effect problems, it's essential to understand the inverse relationship between frequency and wavelength. This relationship is governed by the equation:
\( c = \lambda f \)
where \( c \) is the speed of light in vacuum, \( \lambda \) is the wavelength, and \( f \) is the frequency.
In other words, as wavelength increases, frequency decreases, and vice versa, since the speed of light is a constant \( (c = 3 \times 10^8 \text{m/s}) \).
In practice, to find the frequency when given a wavelength (like in the exercise provided), you would rearrange this equation to:
\( f = \frac{c}{\lambda} \), and then plug in the values for the speed of light and the wavelength. This concept is essential because the energy of a photon depends directly on its frequency, not its wavelength. Therefore, by knowing one (either frequency or wavelength), we can determine the other, and subsequently calculate the energy of the photon, which is a critical step in addressing problems related to the photoelectric effect.