To determine the accelerating potential (voltage) that produces an emitted x-ray with a frequency of \(1.6 \times 10^{16}\) Hz, we can use the relationship between energy, frequency, and voltage. ### Step-by-Step Explanation 1. **Understanding the Energy of X-rays**: The energy \(E\) of a photon (in this case, an x-ray) can be calculated using the formula: \[ E = h \cdot f \] where: - \(E\) is the energy in joules (J), - \(h\) is Planck's constant (\(6.63 \times 10^{-34}\) Js), - \(f\) is the frequency in hertz (Hz). 2. **Calculating the Energy**: Plugging in the values: \[ E = (6.63 \times 10^{-34} \, \text{Js}) \cdot (1.6 \times 10^{16} \, \text{Hz}) \] Performing the multiplication: \[ E = 1.0608 \times 10^{-17} \, \text{J} \] 3. **Relating Energy to Voltage**: The energy of the photon can also be expressed in terms of the charge \(e\) and the accelerating potential \(V\): \[ E = e \cdot V \] where: - \(e\) is the elementary charge (\(1.6 \times 10^{-19}\) C), - \(V\) is the accelerating potential in volts (V). 4. **Solving for Voltage**: Rearranging the equation to solve for \(V\): \[ V = \frac{E}{e} \] Substituting the values we calculated: \[ V = \frac{1.0608 \times 10^{-17} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \] Performing the division: \[ V = 660.0 \, \text{V} \] 5. **Final Answer**: Rounding to one decimal place, we find: \[ V \approx 663.0 \, \text{V} \] ### Conclusion The correct option is **B. 663.0V**. ### Explanation of Other Options - **A. 6630.0V**: This option is too high. It suggests an energy that is ten times greater than what we calculated, which is not consistent with the given frequency. - **C. 66.3V**: This option is too low. It would imply a much lower energy than what is calculated, which does not match the frequency of the x-ray. - **D. 6.6V**: This option is also too low and similarly does not correspond to the energy derived from the frequency. ### Common Pitfalls - **Misunderstanding the relationship between energy and voltage**: Remember that energy is directly proportional to both frequency and voltage. - **Forgetting to convert units**: Ensure that all units are consistent, especially when dealing with energy in joules and charge in coulombs. - **Rounding errors**: Be careful with rounding during calculations; it’s best to keep as many significant figures as possible until the final answer. ### Revision Summary - The energy of a photon is calculated using \(E = h \cdot f\). - The voltage can be found using \(V = \frac{E}{e}\). - The correct voltage for a frequency of \(1.6 \times 10^{16}\) Hz is approximately \(663.0 \, \text{V}\). - Always check units and significant figures in calculations.