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Understanding wavelength conversion is a fundamental step in many physics problems involving light. Wavelength is typically measured in nanometers (nm) when dealing with visible light, as it is in the original exercise where we have a wavelength of 488 nm. However, to perform calculations involving speed and frequency, it is necessary to convert this measurement into meters. This is because the speed of light, denoted as \( c \), is given in meters per second (\( m/s \)). The conversion is straightforward: multiply the wavelength in nanometers by \( 10^{-9} \) to convert to meters.

• In our example: \( 488 \text{ nm} = 488 \times 10^{-9} \text{ m} \).

Once you have the wavelength in meters, you can calculate the frequency \( f \) of the light using the equation \( f = \frac{c}{\lambda} \), where \( \lambda \) is the wavelength and \( c \) is approximately \( 3 \times 10^8 \text{ m/s} \). This calculation is crucial because frequency directly relates to photon energy, which is the next step in solving photoelectric effect problems. In this example, the frequency is calculated to be \( 6.15 \times 10^{14} \text{ Hz} \).

Once the frequency \( f \) of light is known, the next step is to determine the energy possessed by each photon of that light. This is done using Planck's formula, \( E = hf \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \). Calculate this energy in Joules first because that's the standard SI unit.

• For our frequency: \( E = (6.626 \times 10^{-34} \text{ J s})(6.15 \times 10^{14} \text{ Hz}) = 4.075 \times 10^{-19} \text{ J} \).

Since problems in photoelectric effect often require energies in electronvolts (eV), you must convert the energy from Joules to eV using the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Thus,

• \( E = \frac{4.075 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 2.54 \text{ eV} \).

Understanding this conversion and calculation is critical as the photon energy will impact the outcome in detecting whether electrons are emitted from a material.

The work function of a material is the minimum energy required to eject an electron from its surface due to the absorption of a photon. This is essential for understanding the photoelectric effect because only photons with energy exceeding this threshold will release electrons. The equation at the heart of this is the photoelectric equation: \( E_{photon} = \text{Work Function (}\phi\text{)} + eV_{stop} \), where \( eV_{stop} \) is the energy equivalent of the stopping potential.

• Given that we have already calculated the photon energy \( E_{photon} \) as 2.54 eV and \( eV_{stop} \) as 0.38 eV (since \( 1 V \cdot e = 1 eV \)), substituting these values in gives: \[ 2.54 = \phi + 0.38 \].

To find the work function \( \phi \), rearrange the formula: \( \phi = 2.54 - 0.38 = 2.16 \text{ eV} \). When considering significant figures and problem options, this rounds to 2.2 eV, matching the correct choice. Understanding how to manipulate this equation and interpret its components helps comprehend how electrons are emitted under various conditions.