Let's break down and answer the question step by step: --- ### **Restatement of the Question** The statement says: > "The kinetic energy of a particle is given by \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is speed. Therefore, a moving electron must have more energy than a photon, since a photon has no mass." Is this reasoning correct? Let's analyze it in detail. --- ## **Step 1: Understanding Kinetic Energy** The formula \(\frac{1}{2}mv^2\) is **classical kinetic energy** for an object with mass \(m\) and speed \(v\). - **Electrons** have mass, so this formula can apply (but only when speeds are much less than the speed of light). - **Photons** have zero rest mass (\(m = 0\)), so plugging into this formula gives zero: \(\frac{1}{2}\times 0 \times v^2 = 0\). But **this formula does not apply to massless particles or to particles moving at or near the speed of light**. --- ## **Step 2: Energy of a Photon** A photon is a quantum of light, a massless particle, but it still carries energy. Its energy is given by: \[ E = h f \] or \[ E = \frac{hc}{\lambda} \] where: - \(h\) = Planck’s constant (\(6.626 \times 10^{-34}\) J·s) - \(f\) = frequency of the photon - \(\lambda\) = wavelength of the photon - \(c\) = speed of light So, **even though a photon has no mass, it has energy**. --- ## **Step 3: Energy of a Moving Electron** For an electron with mass, there are two cases: - **Non-relativistic** (slow): Use \(E_{kin} = \frac{1}{2}mv^2\) - **Relativistic** (fast, close to \(c\)): Use the full energy-momentum relation: \[ E^2 = (mc^2)^2 + (pc)^2 \] where \(p\) is momentum. Total energy: \(E = \gamma mc^2\) Kinetic energy: \(E_{kin} = (\gamma - 1)mc^2\) (\(\gamma = 1/\sqrt{1 - v^2/c^2}\)) --- ## **Step 4: Comparing Electron and Photon Energies** - **Photon**: Can have any energy, depending on its frequency. - **Electron**: Can also have a range of kinetic energies, depending on its speed. *There is **no general rule** that a moving electron must have more energy than a photon.* #### **Example** - A visible light photon (\(\lambda = 500\) nm): \(E = \frac{hc}{\lambda} \approx 4 \times 10^{-19}\) J (\(\approx 2.5\) eV) - A slow-moving electron (e.g., with \(v = 10^4\) m/s): \(E_{kin} = \frac{1}{2} m v^2 \approx \frac{1}{2} \times 9.11 \times 10^{-31} \times (10^4)^2 \approx 4.6 \times 10^{-23}\) J (\(\approx 2.9 \times 10^{-4}\) eV) *This is much less than the photon’s energy!* #### **High-Energy Photon** - A gamma-ray photon can have energies of \(> 1\) MeV (\(> 1.6 \times 10^{-13}\) J), which is much greater than most electrons' kinetic energies. --- ## **Step 5: Conclusion** **The statement is incorrect.** - The formula \(\frac{1}{2}mv^2\) does not describe the energy of a photon. - Photons can have significant energy even without mass, as given by \(E = hf\). - Whether a moving electron has more energy than a photon depends on their respective speeds (for the electron) and frequency (for the photon). - A photon can easily have more energy than a moving electron, or vice versa, depending on the situation. --- ## **Final Answer (Summary)** > **The statement is not correct.** The formula for kinetic energy, \(\frac{1}{2}mv^2\), only applies to objects with mass moving at speeds much less than the speed of light, and does not apply to photons, which are massless. Despite having no mass, photons have energy given by \(E = hf\). Depending on the situation, a photon can have more or less energy than a moving electron; there is no rule that a moving electron must always have more energy than a photon. Thus, the reasoning in the statement is flawed. --- **[25 marks: Award for clear explanation of energy formulas, photon energy, relativistic considerations, numerical example, and logical conclusion.]**

Let's analyze the problem carefully, step by step, providing detailed explanations, relevant formulas, and calculations. --- ## Restatement of the Question The statement under consideration is: > "The kinetic energy of a particle is given by \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is speed. Therefore, a moving electron must have more energy than a photon, since a photon has no mass." **Is this reasoning correct?** --- ## Step 1: Understand the classical kinetic energy formula The classical kinetic energy (KE) formula: \[ KE = \frac{1}{2} m v^2 \] applies **only** to particles with mass moving at speeds much less than the speed of light (\(v \ll c\)). It **does not** apply accurately to particles moving at relativistic speeds or massless particles. --- ## Step 2: Consider the energy of a photon Photons are **massless particles** (\(m=0\)), but they **do carry energy** given by: \[ E = h f = \frac{hc}{\lambda} \] where: - \(h = 6.626 \times 10^{-34}\, \text{Js}\) (Planck's constant) - \(f\) = frequency of the photon - \(c = 3.00 \times 10^8\, \text{m/s}\) (speed of light) - \(\lambda\) = wavelength of the photon **Key point:** photons have **zero rest mass** but **non-zero energy**. --- ## Step 3: Energy of a moving electron (relativistic and non-relativistic) ### Non-relativistic case (\(v \ll c\)): \[ E_{kin} = \frac{1}{2} m v^2 \] ### Relativistic case (\(v \sim c\)): Total energy \(E\) is: \[ E = \gamma mc^2 \] where: \[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \] and the **relativistic kinetic energy**: \[ E_{kin} = (\gamma - 1) mc^2 \] --- ## Step 4: Numerical examples to compare energies ### Example 1: A typical visible light photon - Wavelength: \(\lambda = 500\, \text{nm} = 5 \times 10^{-7}\,\text{m}\) Calculate photon energy: \[ E_{photon} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{5 \times 10^{-7}} \approx 3.97 \times 10^{-19}\, \text{J} \] which is roughly **2.5 eV** (since \(1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J}\)). --- ### Example 2: A slow electron - Electron mass: \(m = 9.11 \times 10^{-31}\, \text{kg}\) - Speed: \(v = 10^4\, \text{m/s}\) Calculate kinetic energy: \[ KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (10^4)^2 \] \[ KE = 0.5 \times 9.11 \times 10^{-31} \times 10^8 = 4.56 \times 10^{-23}\, \text{J} \] which is about **2.85 \times 10^{-4} eV**, **much less** than the photon energy (~2.5 eV). ### Example 3: A high-energy photon (gamma ray) - Wavelength: \(\lambda = 10^{-12}\, \text{m}\) (hard gamma-ray photon) \[ E_{photon} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{10^{-12}} = 1.99 \times 10^{-13}\, \text{J} \] which is about **1.24 MeV** (since \(1\, \text{MeV} \approx 1.602 \times 10^{-13}\, \text{J}\)). - Electron with KE of 1 MeV has: \[ E_{kin} \approx 1\, \text{MeV} = 1.602 \times 10^{-13}\, \text{J} \] which is **less** than the photon energy. --- ## Step 5: Is the statement correct? **Summary of findings:** - The classical kinetic energy formula (\(\frac{1}{2} mv^2\)) **does not** accurately describe the energy of relativistic electrons or massless particles like photons. - **Photons**, despite having no mass, can have **more** energy than a moving electron, depending on their frequency/wavelength. - **Electrons** can have **less** or **more** energy than a photon depending on their speed and the photon’s wavelength. --- ## **Final conclusion:** > **The statement is incorrect.** > - The kinetic energy formula \(\frac{1}{2} mv^2\) applies only to particles with mass moving at speeds much less than \(c\), and **does not** apply to photons. > - Photons, even with zero rest mass, **carry energy** \(E = hf\), which can be **greater or less** than the energy of a moving electron. > - Therefore, **a moving electron does not necessarily have more energy than a photon**; the two energies are comparable only when considering their actual energy formulas, not classical KE. --- ## **Summary:** | Aspect | Explanation | Numerical Example | |------------------------------------------------|----------------------------------------------------------------------------------------|----------------------------------------------------------------| | Classical KE formula | Applies only to massive particles at low speeds | Not valid for photons or relativistic electrons | | Photons’ energy | \(E = hf\), no rest mass, can be very high or low depending on frequency | Visible photon: \(\sim 2.5\, \text{eV}\); gamma photon: \(\sim 1\, \text{MeV}\) | | Electron energy (relativistic) | \(E = \gamma mc^2\); can be much larger than classical KE for high speeds | An electron at high \(\gamma\) can have energy comparable to or greater than photon energy | | **Conclusion** | The reasoning that "a moving electron must have more energy than a photon" is false | Photons can carry more energy than electrons, depending on their wavelength | --- **This completes the detailed explanation and calculations.**