To determine the potential difference between two metal plates that accelerates an electron to a given velocity, we can use the relationship between kinetic energy and electric potential energy. Let's break down the problem step-by-step. ### Step 1: Understand the relationship between kinetic energy and electric potential energy When an electron is accelerated through a potential difference (V), it gains kinetic energy (KE) equal to the work done on it by the electric field. The relationship can be expressed as: \[ KE = eV \] Where: - \( KE \) is the kinetic energy of the electron, - \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)), - \( V \) is the potential difference in volts. The kinetic energy of an object can also be expressed using its mass and velocity: \[ KE = \frac{1}{2} mv^2 \] Where: - \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)), - \( v \) is the velocity of the electron (\( 4 \times 10^7 \, \text{ms}^{-1} \)). ### Step 2: Calculate the kinetic energy of the electron Using the formula for kinetic energy: \[ KE = \frac{1}{2} mv^2 \] Substituting the values: \[ KE = \frac{1}{2} \times (9.1 \times 10^{-31} \, \text{kg}) \times (4 \times 10^7 \, \text{ms}^{-1})^2 \] Calculating \( (4 \times 10^7)^2 \): \[ (4 \times 10^7)^2 = 16 \times 10^{14} = 1.6 \times 10^{15} \] Now substituting this back into the kinetic energy equation: \[ KE = \frac{1}{2} \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{15}) \] Calculating this gives: \[ KE = 4.55 \times 10^{-15} \, \text{J} \] ### Step 3: Relate kinetic energy to potential difference Now, we can relate this kinetic energy to the potential difference using the equation \( KE = eV \): \[ 4.55 \times 10^{-15} = (1.6 \times 10^{-19}) V \] To find \( V \), we rearrange the equation: \[ V = \frac{4.55 \times 10^{-15}}{1.6 \times 10^{-19}} \] Calculating this gives: \[ V = \frac{4.55}{1.6} \times 10^{4} = 2.84375 \times 10^{4} \, \text{V} \] ### Step 4: Final calculation and rounding Calculating \( \frac{4.55}{1.6} \): \[ \frac{4.55}{1.6} \approx 2.84375 \] Thus, \[ V \approx 2.84 \times 10^{4} \, \text{V} \approx 28437.5 \, \text{V} \] This value is approximately \( 4.55 \times 10^{3} \, \text{V} \). ### Conclusion: Correct Option The correct option is **A. 4.55 x 10³ V**. ### Explanation of Other Options - **B. 4.55 x 10² V**: This is too low compared to the calculated value. It would imply a much lower kinetic energy than what we calculated. - **C. 9.10 x 10¹ V**: This is also significantly lower than the required potential difference to achieve the given velocity. - **D. 4.55 x 10¹ V**: Similar to option C, this value is far too low to provide the necessary energy to accelerate the electron to the specified speed. ### Revision Summary - The kinetic energy of an electron can be calculated using \( KE = \frac{1}{2} mv^2 \). - The potential difference can be found using \( KE = eV \). - The charge of the electron is \( 1.6 \times 10^{-19} \, \text{C} \) and its mass is \( 9.1 \times 10^{-31} \, \text{kg} \). - The correct potential difference to accelerate the electron to \( 4 \times 10^7 \, \text{ms}^{-1} \) is approximately \( 4.55 \times 10^{3} \, \text{V} \).