| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we need to use the Einstein's photoelectric equation, which is given by: hf = φ + KE where h is the Planck's constant, f is the frequency of the incident light, φ is the work function of the metal (in this case, lithium), and KE is the kinetic energy of the emitted electrons. We are given the wavelength of the incident light (400 nm), which we can use to find the frequency using the relation: f = c / λ where c is the speed of light and λ is the wavelength. f = (3 x 10^8 m/s) / (400 x 10^-9 m) = 7.5 x 10^14 Hz Now, we can use the Einstein's photoelectric equation to find the kinetic energy of the emitted electrons: hf = φ + KE (6.626 x 10^-34 J s) x (7.5 x 10^14 Hz) = (2.93 eV) x (1.6 x 10^-19 J/eV) + KE KE = 2.04 x 10^-19 J The stopping potential is the potential difference required to stop the emitted electrons from reaching the anode. It can be found using the equation: eV = KE where e is the charge of an electron. V = KE / e = (2.04 x 10^-19 J) / (1.6 x 10^-19 C) = 1.275 V Therefore, the stopping potential is 1.275 V. |
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| deepseekr1 | The stopping potential can be calculated using the photoelectric effect equation. Here's the step-by-step solution: 1. **Photon Energy Calculation**: - Wavelength \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \) - Use \( E_{\text{photon}} = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \, \text{J·s} \) and \( c = 3.00 \times 10^8 \, \text{m/s} \): \[ E_{\text{photon}} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J} \] - Convert to electron volts (\( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \)): \[ E_{\text{photon}} = \frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 3.10 \, \text{eV} \] 2. **Kinetic Energy of Electrons**: - Work function \( \phi = 2.93 \, \text{eV} \) - Maximum kinetic energy \( K_{\text{max}} = E_{\text{photon}} - \phi \): \[ K_{\text{max}} = 3.10 \, \text{eV} - 2.93 \, \text{eV} = 0.17 \, \text{eV} \] 3. **Stopping Potential**: - \( K_{\text{max}} = eV_0 \), so \( V_0 = \frac{K_{\text{max}}}{e} \). Since \( K_{\text{max}} \) is in eV, \( V_0 = 0.17 \, \text{V} \). **Answer**: The stopping potential is \(\boxed{0.17 \, \text{V}}\). |
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| gpto1 | To solve this problem, we'll use the photoelectric effect equation, which describes how photons eject electrons from a metal surface when light shines upon it. The key equation is: \[ K_{\text{max}} = h\nu - \phi \] where: - \( K_{\text{max}} \) is the maximum kinetic energy of the ejected electrons. - \( h \) is Planck's constant. - \( \nu \) is the frequency of the incident light. - \( \phi \) is the work function (the minimum energy needed to eject an electron from the metal). The stopping potential \( V_s \) is the voltage needed to stop the most energetic electrons, and it's related to the maximum kinetic energy by: \[ eV_s = K_{\text{max}} \] where \( e \) is the elementary charge (the charge of an electron). **Step 1: Convert Wavelength to Energy** First, we need to find the energy of the incident photons using the wavelength provided. The energy of a photon is given by: \[ E_{\text{photon}} = h\nu = \frac{hc}{\lambda} \] where: - \( c \) is the speed of light (\( 3.00 \times 10^8 \) m/s). - \( \lambda \) is the wavelength of the light. But since we'll be working with electronvolts (eV) and nanometers (nm), it's convenient to use the combined constant \( hc \) in units of eV·nm: \[ hc = 1240 \text{ eV·nm} \] So, the energy of the photons is: \[ E_{\text{photon}} = \frac{1240 \text{ eV·nm}}{\lambda (\text{nm})} \] Given \( \lambda = 400 \text{ nm} \): \[ E_{\text{photon}} = \frac{1240 \text{ eV·nm}}{400 \text{ nm}} = 3.10 \text{ eV} \] **Step 2: Calculate the Maximum Kinetic Energy** Subtract the work function \( \phi \) of lithium from the photon energy to get the maximum kinetic energy of the ejected electrons: \[ K_{\text{max}} = E_{\text{photon}} - \phi = 3.10 \text{ eV} - 2.93 \text{ eV} = 0.17 \text{ eV} \] **Step 3: Find the Stopping Potential** Using the relation between kinetic energy and stopping potential: \[ eV_s = K_{\text{max}} \] Since \( K_{\text{max}} \) is already in electronvolts, and one electronvolt is the energy gained by an electron moving through a potential difference of one volt, the stopping potential is simply: \[ V_s = K_{\text{max}} = 0.17 \text{ V} \] **Answer:** The stopping potential is **0.17 volts (V)**. |
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| GPT-4o | To solve this problem, we need to use the photoelectric equation and the concept of stopping potential. The stopping potential is the voltage required to stop the most energetic photoelectrons emitted from the material. Here's a step-by-step guide to solving this problem: 1. **Convert the wavelength of the incident light into energy:** The energy \( E \) of a photon can be calculated using the equation: \[ E = \frac{hc}{\lambda} \] where: - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)) - \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)) - \( \lambda \) is the wavelength of the incident light (400 nm or \( 400 \times 10^{-9} \, \text{m} \)) Plugging in the values: \[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) (3.00 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.9878 \times 10^{-25} \, \text{J}}{400 \times 10^{-9}} \] \[ E = 4.9695 \times 10^{-19} \, \text{J} \] 2. **Convert the energy from joules to electron volts (eV):** 1 eV = \( 1.602 \times 10^{-19} \) J. \[ E = \frac{4.9695 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \] \[ E \approx 3.10 \, \text{eV} \] 3. **Apply the photoelectric equation:** The photoelectric equation is given by: \[ E_{\text{photon}} = \phi + K_{\text{max}} \] where: - \( E_{\text{photon}} \) is the energy of the incident photon - \( \phi \) is the work function of the material (2.93 eV for lithium) - \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electrons Rearranging for \( K_{\text{max}} \): \[ K_{\text{max}} = E_{\text{photon}} - \phi \] \[ K_{\text{max}} = 3.10 \, \text{eV} - 2.93 \, \text{eV} \] \[ K_{\text{max}} = 0.17 \, \text{eV} \] 4. **Relate the maximum kinetic energy to the stopping potential:** The stopping potential \( V_s \) is related to the maximum kinetic energy by the equation: \[ K_{\text{max}} = eV_s \] where \( e \) is the elementary charge (1 eV). Therefore, the stopping potential \( V_s \) is: \[ V_s = K_{\text{max}} / e \] \[ V_s = 0.17 \, \text{V} \] So, the stopping potential for the given problem is \( \boxed{0.17 \, \text{V}} \). |
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| llama3.1-70B-Instruct | A classic problem in physics! To calculate the stopping potential, we need to use the photoelectric effect equation: K = hf - φ where: K = kinetic energy of the electron (in eV) h = Planck's constant (6.626 x 10^-34 J s) f = frequency of the light (in Hz) φ = work function of the material (in eV) First, we need to find the frequency of the light: f = c / λ where: c = speed of light (3 x 10^8 m/s) λ = wavelength of the light (400 nm = 4 x 10^-7 m) f = (3 x 10^8 m/s) / (4 x 10^-7 m) = 7.5 x 10^14 Hz Now, we can plug in the values: K = hf - φ = (6.626 x 10^-34 J s) x (7.5 x 10^14 Hz) - 2.93 eV = 4.97 eV - 2.93 eV = 2.04 eV The stopping potential (V) is equal to the kinetic energy (K) in eV: V = K = 2.04 eV Therefore, the stopping potential is 2.04 V. |
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| llama3.1-405B-Instruct | To calculate the stopping potential, we can use the formula: Stopping potential (V) = (hc/λ) - φ where: h = Planck's constant (6.626 x 10^-34 J s) c = speed of light (3 x 10^8 m/s) λ = wavelength of light (400 nm = 400 x 10^-9 m) φ = work function of lithium (2.93 eV) First, let's convert the work function from eV to J: φ = 2.93 eV x (1.602 x 10^-19 J/eV) = 4.69 x 10^-19 J Now, we can plug in the values: Stopping potential (V) = (hc/λ) - φ = (6.626 x 10^-34 J s x 3 x 10^8 m/s) / (400 x 10^-9 m) - 4.69 x 10^-19 J = 4.97 x 10^-19 J - 4.69 x 10^-19 J = 0.28 x 10^-19 J To convert this to volts, we can divide by the elementary charge (1.602 x 10^-19 C): Stopping potential (V) = 0.28 x 10^-19 J / (1.602 x 10^-19 C) = 0.175 V Therefore, the stopping potential is approximately 0.175 V. |
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| DeepSeek-Coder-V2 | To solve the problem of calculating the stopping potential for light of wavelength 400 nm incident upon lithium with a work function of 2.93 eV, we can follow these steps: 1. **Convert the wavelength to frequency**: The relationship between wavelength and frequency is given by: \[ \lambda = \frac{c}{\nu} \] where \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s), \(\lambda\) is the wavelength, and \(\nu\) is the frequency. Given \(\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\): \[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{400 \times 10^{-9} \, \text{m}} = 7.5 \times 10^{14} \, \text{Hz} \] 2. **Calculate the energy of the photon**: The energy of a photon is given by Planck's equation: \[ E = h \nu \] where \( h \) is Planck's constant (approximately \( 6.626 \times 10^{-34} \, \text{Js} \)). \[ E = (6.626 \times 10^{-34} \, \text{Js}) \times (7.5 \times 10^{14} \, \text{Hz}) = 4.9695 \times 10^{-19} \, \text{J} \] 3. **Convert the energy from joules to electron volts**: \[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \] \[ E = \frac{4.9695 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 3.102 \, \text{eV} \] 4. **Calculate the stopping potential**: The stopping potential \( V_0 \) is given by the equation: \[ eV_0 = E - \phi \] where \( e \) is the elementary charge (approximately \( 1.602 \times 10^{-19} \, \text{C} \)), \( E \) is the energy of the photon, and \( \phi \) is the work function of lithium. \[ V_0 = \frac{E - \phi}{e} = \frac{3.102 \, \text{eV} - 2.93 \, \text{eV}}{1 \, \text{eV}} = \frac{0.172 \, \text{eV}}{1 \, \text{eV}} = 0.172 \, \text{V} \] Therefore, the stopping potential is: \[ \boxed{0.172} \] |
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| Qwen2-72B-Instruct | To solve the problem, we need to use the Einstein's photoelectric equation, which is given by: hf = phi + KE where h is the Planck's constant, f is the frequency of the incident light, phi is the work function of the metal (in this case, lithium), and KE is the kinetic energy of the emitted electrons. We are given the wavelength of the incident light (400 nm), which we can use to find the frequency using the relation: f = c / lambda where c is the speed of light and lambda is the wavelength. f = (3 x 10^8 m/s) / (400 x 10^-9 m) = 7.5 x 10^14 Hz Now, we can use the Einstein's photoelectric equation to find the kinetic energy of the emitted electrons: hf = phi + KE (6.626 x 10^-34 J s) x (7.5 x 10^14 Hz) = (2.93 eV) x (1.6 x 10^-19 J/eV) + KE KE = 2.04 x 10^-19 J The stopping potential is the potential difference required to stop the emitted electrons from reaching the anode. It can be found using the equation: eV = KE where e is the charge of an electron. V = KE / e = (2.04 x 10^-19 J) / (1.6 x 10^-19 C) = 1.275 V Therefore, the stopping potential is 1.275 V. |
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| baichuan4 | To calculate the stopping potential, we'll use the photoelectric effect equation: E = hc / λ - φ where: - E is the maximum kinetic energy of the ejected electron (in eV), - h is Planck's constant (4.135667696 × 10^-15 eV s), - c is the speed of light (299,792,458 m/s), - λ is the wavelength of the incident light (400 nm), - φ is the work function of the material (2.93 eV for lithium). First, let's convert the wavelength to meters: λ = 400 nm = 400 × 10^-9 m Now, let's calculate the energy of the incident light: E_light = hc / λ = (4.135667696 × 10^-15 eV s) × (299,792,458 m/s) / (400 × 10^-9 m) ≈ 3.11 eV Next, we subtract the work function to find the maximum kinetic energy of the ejected electrons: E_k = E_light - φ = 3.11 eV - 2.93 eV = 0.18 eV The stopping potential (V_0) is the potential difference that stops the electrons from moving. It is equal to the maximum kinetic energy of the ejected electrons divided by the electron charge (1.60217662 × 10^-19 C): V_0 = E_k / e = 0.18 eV / (1.60217662 × 10^-19 C) ≈ 0.1123 V Therefore, the stopping potential for the given problem is approximately 0.1123 volts. |
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| doubao | The energy of a photon is given by E = hf = hc/λ, where h is Planck's constant (6.626 × 10^-34 Js), c is the speed of light (3 × 10^8 m/s), and λ is the wavelength. Converting the wavelength to meters: 400 nm = 400 × 10^-9 m. The energy of the photon is E = (6.626 × 10^-34 Js × 3 × 10^8 m/s) / (400 × 10^-9 m) = 4.9695 × 10^-19 J. To convert this to electron volts, divide by the charge of an electron (1.602 × 10^-19 C), so the energy is (4.9695 × 10^-19 J) / (1.602 × 10^-19 C) = 3.1 eV. The stopping potential is equal to the energy of the photon minus the work function. So the stopping potential is 3.1 eV - 2.93 eV = 0.17 eV. Converting to volts: 0.17 eV = 0.17 V. |
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| minimax | To calculate the stopping potential for electrons emitted from lithium when light of wavelength 400 nm is incident upon it, we can use the photoelectric effect equation: \[ E_k = h\nu - \phi \] where: - \( E_k \) is the maximum kinetic energy of the emitted electrons, - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J·s), - \( \nu \) is the frequency of the incident light, - \( \phi \) is the work function of lithium (2.93 eV in this case). First, we need to convert the wavelength of the incident light to frequency. The relationship between wavelength (\( \lambda \)) and frequency (\( \nu \)) is given by: \[ c = \lambda \nu \] where \( c \) is the speed of light in a vacuum (\( 3 \times 10^8 \) m/s). Given the wavelength \( \lambda = 400 \) nm, we convert it to meters: \[ \lambda = 400 \times 10^{-9} \text{ m} \] Now we can calculate the frequency: \[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} \] \[ \nu = \frac{3 \times 10^8}{400 \times 10^{-9}} \] \[ \nu = \frac{3}{4} \times 10^{15} \text{ Hz} \] Next, we need to calculate the energy of one photon of this light, which is given by: \[ E = h\nu \] \[ E = (6.626 \times 10^{-34} \text{ J·s}) \times \left(\frac{3}{4} \times 10^{15} \text{ Hz}\right) \] \[ E = 6.626 \times 10^{-34} \times \frac{3}{4} \times 10^{15} \] \[ E = 4.9695 \times 10^{-19} \text{ J} \] Since the work function \( \phi \) is given in electron volts (eV), we need to convert the energy of the photon from joules to eV. The conversion factor is: \[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \] So the energy of the photon in eV is: \[ E = \frac{4.9695 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \] \[ E \approx 3.10 \text{ eV} \] Now we can find the maximum kinetic energy of the emitted electrons: \[ E_k = E - \phi \] \[ E_k = 3.10 \text{ eV} - 2.93 \text{ eV} \] \[ E_k = 0.17 \text{ eV} \] The stopping potential (\( V_s \)) is the potential that would need to be applied to stop the most energetic electrons. Since kinetic energy is equal to the charge of an electron (\( e = 1.602 \times 10^{-19} \text{ C} \)) times the stopping potential (\( V_s \)), we have: \[ E_k = e V_s \] \[ V_s = \frac{E_k}{e} \] \[ V_s = \frac{0.17 \text{ eV}}{1.602 \times 10^{-19} \text{ C}} \] \[ V_s \approx \frac{0.17}{1.602} \times 10^{19} \] \[ V_s \approx 1.06 \times 10^{18} \text{ V} \] \[ V_s \approx 1.06 \text{ V} \] Therefore, the stopping potential for electrons emitted from lithium when light of wavelength 400 nm is incident upon it is approximately 1.06 V. |
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| kimi | To calculate the stopping potential, we first need to determine if the incident light has enough energy to eject electrons from the lithium surface. This can be done by comparing the energy of the incident photons to the work function (phi) of lithium. The energy of a photon can be calculated using the formula: \[ E = \frac{hc}{\lambda} \] where: - \( E \) is the energy of the photon, - \( h \) is Planck's constant (\(6.626 \times 10^{-34}\) Js), - \( c \) is the speed of light in a vacuum (\(3 \times 10^8\) m/s), - \( \lambda \) is the wavelength of the light. Given that the wavelength (\( \lambda \)) is 400 nm (which is \( 400 \times 10^{-9} \) m), we can calculate the energy of the photon: \[ E = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} \] \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \] \[ E = \frac{1.9878 \times 10^{-25}}{400 \times 10^{-9}} \] \[ E = 4.9695 \times 10^{-19} \text{ J} \] Now, we convert this energy to electron volts (eV): \[ E_{\text{eV}} = \frac{E}{e} \] \[ E_{\text{eV}} = \frac{4.9695 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \] \[ E_{\text{eV}} \approx 3.1 \text{ eV} \] Since the energy of the incident photons (3.1 eV) is greater than the work function of lithium (2.93 eV), the light will be able to eject electrons from the lithium surface. The stopping potential (\( V_s \)) is the minimum potential that will stop the ejected photoelectrons from reaching the anode. It can be calculated using the equation: \[ eV_s = \phi - E_{\text{photon}} \] Where: - \( eV_s \) is the stopping potential in electron volts, - \( \phi \) is the work function of lithium in electron volts, - \( E_{\text{photon}} \) is the energy of the incident photons in electron volts. \[ V_s = \frac{\phi - E_{\text{photon}}}{e} \] \[ V_s = \frac{2.93 \text{ eV} - 3.1 \text{ eV}}{1.602 \times 10^{-19} \text{ J/eV}} \] \[ V_s = \frac{-0.17 \text{ eV}}{1.602 \times 10^{-19} \text{ J/eV}} \] Since the energy of the photons is greater than the work function, the stopping potential should be negative, indicating that the electrons have enough energy to overcome the work function and reach the anode without stopping. However, if we want to find the potential that would stop the electrons, we take the absolute value: \[ V_s \approx \frac{0.17 \text{ eV}}{1.602 \times 10^{-19} \text{ J/eV}} \] \[ V_s \approx 1.06 \times 10^{19} \text{ V} \] This is a very large value for the stopping potential, which indicates that the electrons would have a very high kinetic energy after being ejected by the photons. In practice, the stopping potential would be much smaller, as it is defined as the potential that stops the most energetic photoelectrons. To find the actual stopping potential, we would use the maximum kinetic energy of the photoelectrons, which is given by: \[ K_{\text{max}} = E_{\text{photon}} - \phi \] \[ K_{\text{max}} = 3.1 \text{ eV} - 2.93 \text{ eV} \] \[ K_{\text{max}} = 0.17 \text{ eV} \] Then, the stopping potential is: \[ eV_s = K_{\text{max}} \ |
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| wenxinyiyan | To solve this problem, we will use the photoelectric effect equation. The photoelectric effect equation relates the energy of the incident photon to the work function of the metal and the maximum kinetic energy of the emitted electron. The equation is: E = φ + KE_max where E is the energy of the incident photon, φ is the work function of the metal, and KE_max is the maximum kinetic energy of the emitted electron. The energy of the incident photon can be calculated using the formula: E = hc/λ where h is Planck's constant, c is the speed of light, and λ is the wavelength of the incident light. The stopping potential, V_stop, is related to the maximum kinetic energy of the emitted electron by the formula: V_stop = KE_max/e where e is the charge of an electron. Now, let's plug in the values given in the problem: λ = 400 nm = 400 × 10^-9 m (convert nm to meters) φ = 2.93 eV = 2.93 × 1.6 × 10^-19 J (convert eV to Joules) h = 6.626 × 10^-34 J·s c = 2.998 × 10^8 m/s e = 1.6 × 10^-19 C First, calculate the energy of the incident photon: E = (6.626 × 10^-34 J·s × 2.998 × 10^8 m/s) / (400 × 10^-9 m) E = 4.966 × 10^-19 J Now, calculate the maximum kinetic energy of the emitted electron: KE_max = E - φ KE_max = 4.966 × 10^-19 J - 2.93 × 1.6 × 10^-19 J KE_max = 1.226 × 10^-19 J Finally, calculate the stopping potential: V_stop = KE_max/e V_stop = (1.226 × 10^-19 J) / (1.6 × 10^-19 C) V_stop = 0.766 V Therefore, the stopping potential is approximately 0.766 V. |
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| zhipu | To calculate the stopping potential for light incident on lithium, we can use the photoelectric equation derived from Einstein's photoelectric effect theory. The equation relates the energy of the incident photons to the work function of the metal and the maximum kinetic energy of the emitted electrons: \[ E_{\text{photon}} = \phi + K_{\text{max}} \] Where: - \( E_{\text{photon}} \) is the energy of the incident photon, - \( \phi \) is the work function of the metal (lithium in this case), - \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electrons. The energy of the photon can be calculated using the equation: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] Where: - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) Js), - \( c \) is the speed of light in a vacuum (\( 3 \times 10^8 \) m/s), - \( \lambda \) is the wavelength of the incident light (400 nm or \( 400 \times 10^{-9} \) m). First, we calculate \( E_{\text{photon}} \): \[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \] \[ E_{\text{photon}} = \frac{1.9878 \times 10^{-25}}{400 \times 10^{-9}} \] \[ E_{\text{photon}} = 4.947 \times 10^{-19} \text{ J} \] To convert this energy into electron volts (eV), we use the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \): \[ E_{\text{photon}} = \frac{4.947 \times 10^{-19}}{1.602 \times 10^{-19}} \text{ eV} \] \[ E_{\text{photon}} \approx 3.09 \text{ eV} \] Now, using the photoelectric equation: \[ 3.09 \text{ eV} = 2.93 \text{ eV} + K_{\text{max}} \] Solving for \( K_{\text{max}} \): \[ K_{\text{max}} = 3.09 \text{ eV} - 2.93 \text{ eV} \] \[ K_{\text{max}} = 0.16 \text{ eV} \] The stopping potential \( V_0 \) is the potential needed to stop the most energetic electrons, and it is numerically equal to \( K_{\text{max}} \) in volts: \[ V_0 = K_{\text{max}} \] \[ V_0 = 0.16 \text{ V} \] Therefore, the stopping potential for light of wavelength 400 nm incident upon lithium is 0.16 V. |
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| gpto1 | To solve this problem, we'll use the photoelectric effect equation, which describes how photons eject electrons from a metal surface when light shines upon it. The key equation is: \[ K_{\text{max}} = h\nu - \phi \] where: - \( K_{\text{max}} \) is the maximum kinetic energy of the ejected electrons. - \( h \) is Planck's constant. - \( \nu \) is the frequency of the incident light. - \( \phi \) is the work function (the minimum energy needed to eject an electron from the metal). The stopping potential \( V_s \) is the voltage needed to stop the most energetic electrons, and it's related to the maximum kinetic energy by: \[ eV_s = K_{\text{max}} \] where \( e \) is the elementary charge (the charge of an electron). **Step 1: Convert Wavelength to Energy** First, we need to find the energy of the incident photons using the wavelength provided. The energy of a photon is given by: \[ E_{\text{photon}} = h\nu = \frac{hc}{\lambda} \] where: - \( c \) is the speed of light (\( 3.00 \times 10^8 \) m/s). - \( \lambda \) is the wavelength of the light. But since we'll be working with electronvolts (eV) and nanometers (nm), it's convenient to use the combined constant \( hc \) in units of eV·nm: \[ hc = 1240 \text{ eV·nm} \] So, the energy of the photons is: \[ E_{\text{photon}} = \frac{1240 \text{ eV·nm}}{\lambda (\text{nm})} \] Given \( \lambda = 400 \text{ nm} \): \[ E_{\text{photon}} = \frac{1240 \text{ eV·nm}}{400 \text{ nm}} = 3.10 \text{ eV} \] **Step 2: Calculate the Maximum Kinetic Energy** Subtract the work function \( \phi \) of lithium from the photon energy to get the maximum kinetic energy of the ejected electrons: \[ K_{\text{max}} = E_{\text{photon}} - \phi = 3.10 \text{ eV} - 2.93 \text{ eV} = 0.17 \text{ eV} \] **Step 3: Find the Stopping Potential** Using the relation between kinetic energy and stopping potential: \[ eV_s = K_{\text{max}} \] Since \( K_{\text{max}} \) is already in electronvolts, and one electronvolt is the energy gained by an electron moving through a potential difference of one volt, the stopping potential is simply: \[ V_s = K_{\text{max}} = 0.17 \text{ V} \] **Answer:** The stopping potential is **0.17 volts (V)**. |
0.17 |
| deepseekv3 | To calculate the stopping potential for the given problem, we need to use the photoelectric equation: \[ eV_0 = h\nu - \phi \] Where: - \( e \) is the charge of an electron (\( e = 1.6 \times 10^{-19} \, \text{C} \)), - \( V_0 \) is the stopping potential (in volts), - \( h \) is Planck's constant (\( h = 6.626 \times 10^{-34} \, \text{J·s} \)), - \( \nu \) is the frequency of the incident light, - \( \phi \) is the work function of lithium (\( \phi = 2.93 \, \text{eV} \)). ### Step 1: Convert the wavelength to frequency The wavelength of the light is given as \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \). The frequency \( \nu \) is related to the wavelength by: \[ \nu = \frac{c}{\lambda} \] Where \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)). \[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{400 \times 10^{-9} \, \text{m}} = 7.5 \times 10^{14} \, \text{Hz} \] ### Step 2: Calculate the energy of the incident photons The energy of the photons is given by: \[ E = h\nu \] \[ E = (6.626 \times 10^{-34} \, \text{J·s}) \times (7.5 \times 10^{14} \, \text{Hz}) = 4.9695 \times 10^{-19} \, \text{J} \] Convert this energy to electron volts (eV) using \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \): \[ E = \frac{4.9695 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 3.106 \, \text{eV} \] ### Step 3: Apply the photoelectric equation Substitute the values into the photoelectric equation: \[ eV_0 = h\nu - \phi \] \[ eV_0 = 3.106 \, \text{eV} - 2.93 \, \text{eV} = 0.176 \, \text{eV} \] ### Step 4: Solve for the stopping potential \( V_0 \) Since \( eV_0 = 0.176 \, \text{eV} \), and \( e = 1.6 \times 10^{-19} \, \text{C} \): \[ V_0 = \frac{0.176 \, \text{eV}}{e} = 0.176 \, \text{V} \] ### Final Answer: The stopping potential \( V_0 \) is: \[ \boxed{0.176 \, \text{V}} \] |
0.17 |