To determine the minimum frequency of light required to eject electrons from a metal surface in the photoelectric effect, we need to use the relationship between the energy of the incoming photons and the work function of the metal. The work function (φ) is the minimum energy needed to remove an electron from the surface of the metal. ### Step-by-Step Explanation 1. **Understanding the Work Function**: The work function (φ) is given in electron volts (eV). In this case, φ = 2.5 eV. This means that to eject an electron from the metal, the incoming light must have a photon energy equal to or greater than 2.5 eV. 2. **Photon Energy and Frequency Relationship**: The energy (E) of a photon is related to its frequency (f) by the equation: \[ E = h \cdot f \] where: - \(E\) is the energy of the photon in joules (J), - \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \, \text{J s}\), - \(f\) is the frequency of the light in hertz (Hz). 3. **Converting Work Function to Joules**: To use the equation, we need to convert the work function from electron volts to joules. The conversion factor is: \[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \] Therefore, the work function in joules is: \[ φ = 2.5 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 4.005 \times 10^{-19} \, \text{J} \] 4. **Calculating Minimum Frequency**: Now we can rearrange the photon energy equation to solve for frequency: \[ f = \frac{E}{h} \] Substituting the values we have: \[ f = \frac{4.005 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J s}} \approx 6.04 \times 10^{14} \, \text{Hz} \] 5. **Choosing the Closest Option**: The calculated frequency of approximately \(6.04 \times 10^{14} \, \text{Hz}\) is closest to option B, which is \(6.3 \times 10^{14} \, \text{Hz}\). Therefore, the correct answer is **B**. ### Why Other Options Are Incorrect - **Option A (5.0 x 10^14 Hz)**: This frequency is lower than the calculated minimum frequency. It would not provide enough energy to overcome the work function of 2.5 eV, hence it cannot eject electrons. - **Option C (4.0 x 10^14 Hz)**: Similar to option A, this frequency is also lower than the required minimum frequency. It does not meet the energy requirement to eject electrons. - **Option D (3.0 x 10^14 Hz)**: This is the lowest frequency option and is far below the calculated minimum frequency. It would definitely not have enough energy to overcome the work function. ### Common Pitfalls - **Forgetting to Convert Units**: Always ensure that energy is in joules when using the formula \(E = h \cdot f\). - **Misunderstanding Work Function**: Remember that the work function is the minimum energy required; any frequency that results in energy below this will not eject electrons. ### Revision Summary - The work function is the minimum energy needed to eject electrons from a metal surface. - The energy of a photon is given by \(E = h \cdot f\). - Convert the work function from eV to joules for calculations. - The minimum frequency can be calculated using \(f = \frac{E}{h}\). - The correct answer for the minimum frequency required to eject electrons from the metal is **B (6.3 x 10^14 Hz)**.