| 模型 | 模型答案 | 正确答案 |
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| 金声玉亮2.0 | To calculate the speed of the electrons after being accelerated through a potential difference of 25,000 volts, we can use the conservation of energy principle. The potential energy of the electrons is converted into kinetic energy as they are accelerated. The kinetic energy (KE) of an electron can be calculated using the formula: KE = 0.5 * m * v^2 where m is the mass of the electron (9.11 × 10^-31 kg), and v is the velocity of the electron in meters per second (m/s). The potential energy (PE) of the electrons can be calculated using the formula: PE = q * V where q is the charge of the electron (1.602 × 10^-19 C), and V is the potential difference (25,000 V). Since the potential energy is converted into kinetic energy, we can equate the two: PE = KE Substituting the values of PE and KE, we get: q * V = 0.5 * m * v^2 Rearranging the equation to solve for v, we get: v = sqrt((2 * q * V) / m) Substituting the values of q, V, and m, we get: v = sqrt((2 * 1.602 × 10^-19 C * 25,000 V) / (9.11 × 10^-31 kg)) v = sqrt((8.01 × 10^-15 J) / (9.11 × 10^-31 kg)) v = sqrt(8.79 × 10^15 m^2/s^2) v = 2.96 × 10^8 m/s Therefore, the speed of the electrons after being accelerated through a potential difference of 25,000 volts is approximately 2.96 × 10^8 m/s. However, this answer is not physically possible as it exceeds the speed of light (2.998 × 10^8 m/s). This discrepancy arises because the relativistic effects become significant at high speeds, and the classical equations used here are not valid. A more accurate calculation would require the use of relativistic equations. |
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| deepseekr1 | The speed of the electrons can be calculated using the principle of energy conservation, where the kinetic energy gained by the electron equals the work done by the electric field. The formula used is: \[ v = \sqrt{\frac{2qV}{m}} \] **Given values:** - Charge of electron (\(q\)) = \(1.6 \times 10^{-19} \, \text{C}\) - Voltage (\(V\)) = 25,000 V - Mass of electron (\(m\)) = \(9.11 \times 10^{-31} \, \text{kg}\) **Calculation:** 1. Compute the numerator: \(2qV = 2 \times 1.6 \times 10^{-19} \times 25,000 = 8 \times 10^{-15} \, \text{J}\) 2. Divide by mass: \(\frac{8 \times 10^{-15}}{9.11 \times 10^{-31}} \approx 8.79 \times 10^{15} \, \text{m}^2/\text{s}^2\) 3. Take the square root: \(v \approx \sqrt{8.79 \times 10^{15}} \approx 9.37 \times 10^{7} \, \text{m/s}\) Considering significant figures (25,000 V has 2 significant figures), the speed is: \[ \boxed{9.4 \times 10^7 \, \text{m/s}} \] |
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| gpto1 | To find the speed of electrons accelerated from rest through a potential difference of 25,000 volts, we need to consider the energy gained by the electrons due to the electric potential. The kinetic energy (KE) gained by each electron is equal to the work done on it by the electric field, which is the product of the electron's charge \(e\) and the potential difference \(V\): \[ \text{Kinetic Energy} = eV \] However, since the kinetic energy is becoming a significant fraction of the electron's rest energy (\(mc^2\)), we need to account for relativistic effects for accurate results. We'll proceed using relativistic mechanics. **Step 1: Calculate the kinetic energy (KE) gained by the electron** The charge of an electron is: \[ e = 1.602 \times 10^{-19} \text{ C} \] The potential difference is: \[ V = 25,000 \text{ V} \] The kinetic energy gained by the electron is: \[ \text{KE} = eV = (1.602 \times 10^{-19} \text{ C})(25,000 \text{ V}) = 4.005 \times 10^{-15} \text{ J} \] **Step 2: Calculate the rest energy (\(mc^2\)) of the electron** The rest mass of an electron is: \[ m = 9.109 \times 10^{-31} \text{ kg} \] The speed of light is: \[ c = 3.00 \times 10^{8} \text{ m/s} \] The rest energy is: \[ E_0 = mc^2 = (9.109 \times 10^{-31} \text{ kg})(3.00 \times 10^{8} \text{ m/s})^2 = 8.187 \times 10^{-14} \text{ J} \] **Step 3: Calculate the Lorentz factor \(\gamma\)** Relativistically, the kinetic energy is related to the rest energy and the Lorentz factor by: \[ \text{KE} = (\gamma - 1) mc^2 \] Rearranged to solve for \(\gamma\): \[ \gamma = 1 + \frac{\text{KE}}{mc^2} = 1 + \frac{4.005 \times 10^{-15} \text{ J}}{8.187 \times 10^{-14} \text{ J}} = 1 + 0.0489 = 1.0489 \] **Step 4: Calculate the speed \(v\) using \(\gamma\)** The Lorentz factor is also related to speed by: \[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \] Rearranged to solve for \(v\): \[ \frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}} \] Plugging in the calculated \(\gamma\): \[ \frac{v}{c} = \sqrt{1 - \frac{1}{(1.0489)^2}} = \sqrt{1 - \frac{1}{1.1001}} = \sqrt{1 - 0.9089} = \sqrt{0.0911} = 0.302 \] Therefore, the speed \(v\) is: \[ v = 0.302c = 0.302 \times (3.00 \times 10^{8} \text{ m/s}) = 9.06 \times 10^{7} \text{ m/s} \] **Answer:** Approximately **\(9.06 \times 10^7\) meters per second**; that's the speed of the electrons after acceleration. |
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| GPT-4o | To solve the problem, we need to determine the speed of electrons that are accelerated through a given potential difference. We can do this using principles of energy conservation and the relationship between kinetic energy and electric potential energy. 1. **Potential Energy and Kinetic Energy**: When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. The work done is given by: \[ W = eV \] where \( e \) is the charge of the electron (\( e \approx 1.602 \times 10^{-19} \) coulombs) and \( V \) is the potential difference (25,000 volts or 25,000 joules per coulomb). 2. **Kinetic Energy**: The kinetic energy (KE) gained by the electron is: \[ KE = \frac{1}{2} mv^2 \] where \( m \) is the mass of the electron (\( m \approx 9.109 \times 10^{-31} \) kg) and \( v \) is the speed of the electron, which we need to find. 3. **Equating the Energies**: Set the kinetic energy equal to the work done by the electric field: \[ eV = \frac{1}{2} mv^2 \] 4. **Solving for \( v \)**: Rearrange the equation to solve for \( v \): \[ v = \sqrt{\frac{2eV}{m}} \] 5. **Substitute the Given Values**: \[ v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} \, \text{C} \times 25,000 \, \text{V}}{9.109 \times 10^{-31} \, \text{kg}}} \] 6. **Calculate the Speed**: \[ v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} \times 25,000}{9.109 \times 10^{-31}}} \] \[ v = \sqrt{\frac{8.01 \times 10^{-15}}{9.109 \times 10^{-31}}} \] \[ v = \sqrt{8.79 \times 10^{15}} \] \[ v \approx 2.97 \times 10^7 \, \text{m/s} \] Thus, the speed of the electrons after being accelerated through a potential difference of 25,000 volts is approximately \( 2.97 \times 10^7 \) meters per second. |
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| llama3.1-70B-Instruct | To calculate the speed of the electrons, we can use the equation for kinetic energy: Kinetic Energy (KE) = (1/2)mv^2 where m is the mass of the electron and v is its speed. The kinetic energy gained by the electron as it accelerates through the potential difference is equal to the energy gained from the electric field, which is given by: KE = eV where e is the charge of the electron (1.6 x 10^-19 C) and V is the potential difference (25,000 V). Equating the two expressions for kinetic energy, we get: (1/2)mv^2 = eV Rearranging to solve for v, we get: v = √(2eV/m) Plugging in the values, we get: v = √(2 x 1.6 x 10^-19 C x 25,000 V / 9.11 x 10^-31 kg) = √(8.8 x 10^14 m^2/s^2) = 2.96 x 10^7 m/s Therefore, the speed of the electrons is approximately 2.96 x 10^7 m/s. |
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| llama3.1-405B-Instruct | To solve this problem, we can use the equation for the kinetic energy of an electron accelerated through a potential difference: K = eV where K is the kinetic energy of the electron, e is the elementary charge (1.602 x 10^-19 C), and V is the potential difference (25,000 V). First, we need to convert the potential difference from volts to joules: V = 25,000 V = 25,000 J/C Now, we can plug in the values: K = eV = (1.602 x 10^-19 C) x (25,000 J/C) = 4.005 x 10^-15 J Next, we can use the equation for kinetic energy in terms of velocity: K = (1/2)mv^2 where m is the mass of the electron (9.109 x 10^-31 kg). Rearranging the equation to solve for velocity, we get: v = sqrt(2K/m) Plugging in the values, we get: v = sqrt(2 x 4.005 x 10^-15 J / 9.109 x 10^-31 kg) = 2.19 x 10^8 m/s Therefore, the speed of the electrons is approximately 2.19 x 10^8 m/s. |
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| DeepSeek-Coder-V2 | To solve the problem of calculating the speed of electrons accelerated through a potential difference of 25,000 volts, we can follow these steps: 1. **Determine the kinetic energy of the electrons:** The kinetic energy \( K \) gained by the electrons is equal to the potential energy \( U \) they lose. The potential energy is given by the product of the charge of the electron \( e \) and the potential difference \( V \): \[ K = U = e \cdot V \] Here, \( e \) is the elementary charge, which is approximately \( 1.602 \times 10^{-19} \) Coulombs, and \( V \) is the potential difference of 25,000 volts. 2. **Calculate the kinetic energy:** \[ K = (1.602 \times 10^{-19} \, \text{C}) \times (25,000 \, \text{V}) = 4.005 \times 10^{-15} \, \text{J} \] 3. **Use the kinetic energy to find the speed of the electrons:** The kinetic energy of a particle with mass \( m \) and speed \( v \) is given by: \[ K = \frac{1}{2} m v^2 \] Solving for \( v \): \[ v = \sqrt{\frac{2K}{m}} \] The mass of an electron \( m \) is approximately \( 9.109 \times 10^{-31} \, \text{kg} \). 4. **Substitute the values and calculate the speed:** \[ v = \sqrt{\frac{2 \times 4.005 \times 10^{-15} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} \] \[ v = \sqrt{\frac{8.01 \times 10^{-15} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} \] \[ v = \sqrt{8.79 \times 10^{15} \, \text{m}^2/\text{s}^2} \] \[ v \approx 2.96 \times 10^8 \, \text{m/s} \] Therefore, the speed of the electrons is approximately \( 2.96 \times 10^8 \, \text{m/s} \). |
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| Qwen2-72B-Instruct | To calculate the speed of the electrons after being accelerated through a potential difference of 25,000 volts, we can use the conservation of energy principle. The potential energy of the electrons is converted into kinetic energy as they are accelerated. The kinetic energy (KE) of an electron can be calculated using the formula: KE = 0.5 * m * v^2 where m is the mass of the electron (9.11 × 10^-31 kg), and v is the velocity of the electron in meters per second (m/s). The potential energy (PE) of the electrons can be calculated using the formula: PE = q * V where q is the charge of the electron (1.602 × 10^-19 C), and V is the potential difference (25,000 V). Since the potential energy is converted into kinetic energy, we can equate the two: PE = KE Substituting the values of PE and KE, we get: q * V = 0.5 * m * v^2 Rearranging the equation to solve for v, we get: v = sqrt((2 * q * V) / m) Substituting the values of q, V, and m, we get: v = sqrt((2 * 1.602 × 10^-19 C * 25,000 V) / (9.11 × 10^-31 kg)) v = sqrt((8.01 × 10^-15 J) / (9.11 × 10^-31 kg)) v = sqrt(8.79 × 10^15 m^2/s^2) v = 2.96 × 10^8 m/s Therefore, the speed of the electrons after being accelerated through a potential difference of 25,000 volts is approximately 2.96 × 10^8 m/s. However, this answer is not physically possible as it exceeds the speed of light (2.998 × 10^8 m/s). This discrepancy arises because the relativistic effects become significant at high speeds, and the classical equations used here are not valid. A more accurate calculation would require the use of relativistic equations. |
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| baichuan4 | To solve this problem, we can use the relationship between the kinetic energy of an electron and the potential energy it gains when accelerated through a potential difference. The equation for the kinetic energy (KE) of an electron is: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass of the electron and \( v \) is its speed. The potential energy (PE) gained by the electron when accelerated through a potential difference \( V \) is: \[ PE = eV \] where \( e \) is the charge of the electron and \( V \) is the potential difference in volts. At the moment the electron strikes the metal target, all of its potential energy has been converted into kinetic energy. Therefore, we can set these two equations equal to each other: \[ \frac{1}{2}mv^2 = eV \] Now, we can rearrange this equation to solve for \( v \): \[ v^2 = \frac{2eV}{m} \] \[ v = \sqrt{\frac{2eV}{m}} \] Let's plug in the values for \( e \) (the charge of an electron is approximately \( 1.6 \times 10^{-19} \) coulombs), \( V \) (25,000 volts), and \( m \) (the mass of an electron is approximately \( 9.11 \times 10^{-31} \) kilograms): \[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19}) \times 25,000}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 25,000}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{\frac{8 \times 10^{-15}}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{8.78 \times 10^{15}} \] \[ v = 9.38 \times 10^7 \, \text{m/s} \] So, the speed of the electrons accelerated through a potential difference of 25,000 volts is approximately \( 9.38 \times 10^7 \) m/s. |
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| doubao | The energy gained by an electron when accelerated through a potential difference V is eV, where e is the elementary charge (1.602 × 10^-19 C). The kinetic energy of the electron is also given by (1/2)mv^2. The potential difference is 25,000 volts = 25000 × 1.602 × 10^-19 J. So, (1/2)mv^2 = 25000 × 1.602 × 10^-19. The mass of an electron is 9.10938356 × 10^-31 kg. (1/2) × 9.10938356 × 10^-31 × v^2 = 25000 × 1.602 × 10^-19 v^2 = (2 × 25000 × 1.602 × 10^-19) / (9.10938356 × 10^-31) v = √[(2 × 25000 × 1.602 × 10^-19) / (9.10938356 × 10^-31)] ≈ 4.19 × 10^7 m/s. |
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| minimax | To calculate the speed of the electrons after being accelerated through a potential difference of 25,000 volts, we can use the principle of conservation of energy. The potential energy that the electrons gain as they are accelerated through the potential difference is converted into kinetic energy. The potential energy (PE) gained by the electron can be calculated using the equation: \[ PE = q \cdot V \] where \( q \) is the charge of the electron and \( V \) is the potential difference. The charge of an electron \( q \) is approximately \( -1.602 \times 10^{-19} \) coulombs, and the potential difference \( V \) is given as 25,000 volts. \[ PE = (-1.602 \times 10^{-19} \text{ C}) \cdot (25,000 \text{ V}) \] \[ PE = -4.005 \times 10^{-15} \text{ J} \] The negative sign indicates that the electron is gaining energy as it is being accelerated towards the positive terminal, but since we are interested in the magnitude of the energy, we can ignore the sign. The kinetic energy (KE) gained by the electron is equal to the potential energy it had before acceleration: \[ KE = PE \] The kinetic energy of an object is also given by the equation: \[ KE = \frac{1}{2} m v^2 \] where \( m \) is the mass of the electron and \( v \) is its speed. The mass of an electron \( m \) is approximately \( 9.109 \times 10^{-31} \) kilograms. Setting the two expressions for kinetic energy equal to each other, we get: \[ \frac{1}{2} m v^2 = PE \] Now we can solve for \( v \): \[ v^2 = \frac{2 \cdot PE}{m} \] \[ v^2 = \frac{2 \cdot (-4.005 \times 10^{-15} \text{ J})}{9.109 \times 10^{-31} \text{ kg}} \] \[ v^2 = \frac{-8.01 \times 10^{-15} \text{ J}}{9.109 \times 10^{-31} \text{ kg}} \] \[ v^2 = -8.79 \times 10^{15} \text{ m}^2/\text{s}^2 \] Taking the square root of both sides to solve for \( v \): \[ v = \sqrt{-8.79 \times 10^{15} \text{ m}^2/\text{s}^2} \] Since speed cannot be negative, we take the absolute value of the square root: \[ v = \sqrt{8.79 \times 10^{15}} \text{ m/s} \] \[ v \approx 2.965 \times 10^7 \text{ m/s} \] Therefore, the speed of the electrons after being accelerated through a potential difference of 25,000 volts is approximately \( 2.965 \times 10^7 \) meters per second. |
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| kimi | To calculate the speed of the electrons after being accelerated through a potential difference of 25,000 volts, we can use the kinetic energy theorem, which states that the kinetic energy gained by the electrons is equal to the work done on them by the electric field. The work done on an electron by the electric field is given by: \[ W = qV \] where \( q \) is the charge of an electron and \( V \) is the potential difference. The charge of an electron is approximately \( 1.6 \times 10^{-19} \) coulombs, and the potential difference \( V \) is 25,000 volts. The kinetic energy \( KE \) of the electron can be calculated as: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass of an electron and \( v \) is the velocity of the electron. Since the work done on the electron is equal to its kinetic energy, we can equate the two: \[ qV = \frac{1}{2}mv^2 \] We can rearrange this to solve for \( v \): \[ v = \sqrt{\frac{2qV}{m}} \] The mass of an electron \( m \) is approximately \( 9.11 \times 10^{-31} \) kilograms. Substituting the values, we get: \[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 25000}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{\frac{8 \times 10^{-18}}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{8.78 \times 10^{12}} \] \[ v \approx 2.96 \times 10^6 \text{ m/s} \] So, the speed of the electrons after being accelerated through a potential difference of 25,000 volts is approximately \( 2.96 \times 10^6 \) meters per second. |
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| wenxinyiyan | To calculate the speed of the electrons, we can use the conservation of energy principle. The electrical potential energy is converted into kinetic energy as the electrons are accelerated through the potential difference. The potential energy of an electron accelerated through a potential difference V is given by: PE = eV where e is the charge of an electron (1.602 x 10^-19 Coulombs) and V is the potential difference (25,000 volts). The kinetic energy of an electron with mass m and speed v is given by: KE = 1/2 mv^2 Since the potential energy is converted into kinetic energy, we can set these two equal to each other: eV = 1/2 mv^2 We can solve this equation for v, the speed of the electron: v = sqrt(2eV/m) Plugging in the values for e (1.602 x 10^-19 C), V (25,000 V), and m (9.11 x 10^-31 kg), we get: v = sqrt(2 * 1.602 x 10^-19 * 25,000 / 9.11 x 10^-31) v ≈ 2.96 x 10^8 m/s So the speed of the electrons is approximately 2.96 x 10^8 m/s. |
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| zhipu | To calculate the speed of the electrons after being accelerated through a potential difference, we can use the principles of energy conservation. The kinetic energy gained by the electrons is equal to the electrical potential energy they acquire as they are accelerated through the potential difference. The kinetic energy (KE) of an electron is given by: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass of the electron and \( v \) is its velocity. The electrical potential energy (PE) gained by an electron when accelerated through a potential difference \( V \) is given by: \[ PE = eV \] where \( e \) is the charge of the electron and \( V \) is the potential difference. Since the kinetic energy gained is equal to the electrical potential energy: \[ \frac{1}{2}mv^2 = eV \] We can solve for \( v \) (the speed of the electron) as follows: \[ v = \sqrt{\frac{2eV}{m}} \] Now we need to plug in the known values: - The charge of an electron \( e \) is approximately \( 1.602 \times 10^{-19} \) coulombs. - The mass of an electron \( m \) is approximately \( 9.109 \times 10^{-31} \) kilograms. - The potential difference \( V \) is 25,000 volts (or \( 25 \times 10^3 \) volts). Plugging these values into the equation: \[ v = \sqrt{\frac{2 \times (1.602 \times 10^{-19} \, \text{C}) \times (25 \times 10^3 \, \text{V})}{9.109 \times 10^{-31} \, \text{kg}}} \] \[ v = \sqrt{\frac{2 \times 1.602 \times 25 \times 10^{-19+3}}{9.109 \times 10^{-31}}} \] \[ v = \sqrt{\frac{2 \times 1.602 \times 25 \times 10^{-16}}{9.109 \times 10^{-31}}} \] \[ v = \sqrt{\frac{80.1 \times 10^{-16}}{9.109 \times 10^{-31}}} \] \[ v = \sqrt{8.783 \times 10^{15}} \] \[ v \approx 9.37 \times 10^7 \, \text{m/s} \] So, the speed of the electrons after being accelerated through a potential difference of 25,000 volts is approximately \( 9.37 \times 10^7 \) meters per second. This is a reasonable order of magnitude for the speed of electrons in such a setup, being a substantial fraction of the speed of light but not exceeding it, as relativistic effects are not considered in this classical calculation. If the speed were approaching the speed of light, we would need to use relativistic equations instead. |
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| gpto1 | To find the speed of electrons accelerated from rest through a potential difference of 25,000 volts, we need to consider the energy gained by the electrons due to the electric potential. The kinetic energy (KE) gained by each electron is equal to the work done on it by the electric field, which is the product of the electron's charge \(e\) and the potential difference \(V\): \[ \text{Kinetic Energy} = eV \] However, since the kinetic energy is becoming a significant fraction of the electron's rest energy (\(mc^2\)), we need to account for relativistic effects for accurate results. We'll proceed using relativistic mechanics. **Step 1: Calculate the kinetic energy (KE) gained by the electron** The charge of an electron is: \[ e = 1.602 \times 10^{-19} \text{ C} \] The potential difference is: \[ V = 25,000 \text{ V} \] The kinetic energy gained by the electron is: \[ \text{KE} = eV = (1.602 \times 10^{-19} \text{ C})(25,000 \text{ V}) = 4.005 \times 10^{-15} \text{ J} \] **Step 2: Calculate the rest energy (\(mc^2\)) of the electron** The rest mass of an electron is: \[ m = 9.109 \times 10^{-31} \text{ kg} \] The speed of light is: \[ c = 3.00 \times 10^{8} \text{ m/s} \] The rest energy is: \[ E_0 = mc^2 = (9.109 \times 10^{-31} \text{ kg})(3.00 \times 10^{8} \text{ m/s})^2 = 8.187 \times 10^{-14} \text{ J} \] **Step 3: Calculate the Lorentz factor \(\gamma\)** Relativistically, the kinetic energy is related to the rest energy and the Lorentz factor by: \[ \text{KE} = (\gamma - 1) mc^2 \] Rearranged to solve for \(\gamma\): \[ \gamma = 1 + \frac{\text{KE}}{mc^2} = 1 + \frac{4.005 \times 10^{-15} \text{ J}}{8.187 \times 10^{-14} \text{ J}} = 1 + 0.0489 = 1.0489 \] **Step 4: Calculate the speed \(v\) using \(\gamma\)** The Lorentz factor is also related to speed by: \[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \] Rearranged to solve for \(v\): \[ \frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}} \] Plugging in the calculated \(\gamma\): \[ \frac{v}{c} = \sqrt{1 - \frac{1}{(1.0489)^2}} = \sqrt{1 - \frac{1}{1.1001}} = \sqrt{1 - 0.9089} = \sqrt{0.0911} = 0.302 \] Therefore, the speed \(v\) is: \[ v = 0.302c = 0.302 \times (3.00 \times 10^{8} \text{ m/s}) = 9.06 \times 10^{7} \text{ m/s} \] **Answer:** Approximately **\(9.06 \times 10^7\) meters per second**; that's the speed of the electrons after acceleration. |
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| deepseekv3 | To calculate the speed of the electrons accelerated through a potential difference, we can use the relationship between the kinetic energy gained by the electrons and the potential difference. Here's the step-by-step solution: --- ### Step 1: Understand the problem Electrons are accelerated from rest through a potential difference \( V = 25,000 \, \text{V} \). We need to find their final speed. --- ### Step 2: Relate potential difference to kinetic energy The kinetic energy \( K \) gained by an electron is equal to the work done by the electric field, which is given by: \[ K = eV \] where: - \( e = 1.6 \times 10^{-19} \, \text{C} \) (charge of an electron), - \( V = 25,000 \, \text{V} \) (potential difference). Substitute the values: \[ K = (1.6 \times 10^{-19}) \times (25,000) = 4 \times 10^{-15} \, \text{J} \] --- ### Step 3: Relate kinetic energy to speed The kinetic energy of an electron is also given by: \[ K = \frac{1}{2} m v^2 \] where: - \( m = 9.11 \times 10^{-31} \, \text{kg} \) (mass of an electron), - \( v \) is the speed of the electron. Rearrange to solve for \( v \): \[ v = \sqrt{\frac{2K}{m}} \] --- ### Step 4: Substitute values and calculate Substitute \( K = 4 \times 10^{-15} \, \text{J} \) and \( m = 9.11 \times 10^{-31} \, \text{kg} \): \[ v = \sqrt{\frac{2 \times (4 \times 10^{-15})}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{\frac{8 \times 10^{-15}}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{8.78 \times 10^{15}} \] \[ v \approx 9.37 \times 10^7 \, \text{m/s} \] --- ### Final Answer: The speed of the electrons is approximately **\( 9.37 \times 10^7 \, \text{m/s} \)**. |
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