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To understand how the photoelectric effect works, we first need to convert the wavelength of light into frequency. Wavelength is usually given in nanometers (nm), and it is a measure of the distance between successive crests of a wave. If an electromagnetic wave has a wavelength of 485 nm, we convert it into meters:
485 nm = 485 x 10^{-9} meters.
Frequency, on the other hand, is the number of wave crests that pass a specific point per second and is measured in Hertz (Hz). To find frequency, use the speed of light equation, which is:

• \(c = \lambda u\)

where:

• \(c\) is the speed of light \(3 \times 10^8\) m/s
• \(\lambda\) is the wavelength
• \(u\) is the frequency

By rearranging the formula to solve for frequency \(u\), we have:

• \(u = \frac{c}{\lambda}\)

Substitute in the values to find that \(u \approx 6.19 \times 10^{14}\) Hz. This step is crucial for calculating the energy of the photons involved.

Planck's equation is vital for finding the energy of photons. In the context of the photoelectric effect, it helps us understand how light can cause electrons to be ejected from a metal surface. The equation is:

• \(E = hu\)

where:

• \(E\) is the energy of the photon
• \(h\) is Planck's constant \(6.626 \times 10^{-34}\) Js
• \(u\) is the frequency of the electromagnetic wave

Substituting the frequency we calculated earlier (\(6.19 \times 10^{14}\) Hz), we find that:

• \(E = 6.626 \times 10^{-34} \times 6.19 \times 10^{14}\)
• \(E \approx 4.10 \times 10^{-19}\) Joules

This tells us each photon has a specific energy based on its wavelength and frequency, which can then be used to calculate the work function.

The work function \(W_0\) is a measure of the minimum energy needed to eject an electron from a material surface. In a photoelectric effect experiment, it's directly related to the energy of the incoming photons.
To find the work function, we need the energy calculated in Joules. We have:

• \(E = 4.10 \times 10^{-19}\) Joules

However, for convenience, this energy is generally converted into electron volts (eV), where:

• 1 eV = \(1.602 \times 10^{-19}\) Joules.

The conversion is performed as follows:

• \(E_{\text{eV}} = \frac{4.10 \times 10^{-19}\ \text{J}}{1.602 \times 10^{-19}\ \text{J/eV}}\)
• \(E_{\text{eV}} \approx 2.56\ \text{eV}\)

This result tells us that it takes 2.56 eV of energy to remove an electron from the surface of the given material.

Photon energy is a central concept in understanding how light interacts with matter in the photoelectric effect. It refers to the energy carried by a single photon, which can be liberated through light absorption to cause effects like electron ejection.
For light to eject electrons, its photon energy must exceed the work function of the material. This means the energy carried by each photon, as calculated by Planck's equation, plays a decisive role.
Key points to consider:

• Photon energy depends on the light's frequency.
• Higher frequency means higher energy photons.
• In our example, photon energy connects directly with the work function calculation, illustrating how photons' energy facilitates the removal of electrons from the metal's surface.

Understanding photon energy hence provides insights into how and why materials respond to different electromagnetic waves based on their frequency and energy.