| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To calculate the photon energy in electronvolts (eV), we use the equation that relates the photon energy (E) to its wavelength (λ) in a vacuum: \[ E = \frac{hc}{\lambda} \] where: - \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} \)), - \( c \) is the speed of light in a vacuum (\( 3.0 \times 10^8 \, \text{m/s} \)), - \( \lambda \) is the wavelength of the light (\( 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)). Substituting the given values: \[ E = \frac{(4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s}) \cdot (3.0 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.2408 \times 10^{-6} \, \text{eV}\cdot\text{m}}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.2408 \times 10^{-6} \, \text{eV}\cdot\text{m}}{4 \times 10^{-7} \, \text{m}} \] \[ E = 3.102 \, \text{eV} \] Thus, the photon energy is approximately 3.102 eV. |
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| deepseekr1 | To calculate the photon energy in eV for light with a wavelength of 400 nm incident on lithium, use the formula: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. A convenient conversion factor is \( hc \approx 1240 \, \text{eV·nm} \). **Steps:** 1. Substitute \( \lambda = 400 \, \text{nm} \) into the formula: \[ E = \frac{1240 \, \text{eV·nm}}{400 \, \text{nm}} = 3.1 \, \text{eV} \] **Answer:** The photon energy is **3.1 eV**. |
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| gpto1 | To calculate the photon energy \( E_{\text{photon}} \) in electron volts (eV) for light of wavelength \( \lambda = 400 \) nanometers (nm), we'll use the fundamental relationship between energy, wavelength, and the constants of physics. ### **Step 1: Understand the Relationship** The energy of a photon is given by the equation: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] Where: - \( E_{\text{photon}} \) is the energy of the photon. - \( h \) is Planck's constant. - \( c \) is the speed of light in a vacuum. - \( \lambda \) is the wavelength of the photon. ### **Step 2: Use Appropriate Units** To ensure consistency, we need to use units that will yield an energy in electron volts (eV). **Important constants:** - **Planck's constant (\( h \)):** - \( h = 6.626 \times 10^{-34} \) Joule\(\cdot\)seconds (J·s) - **Speed of light (\( c \)):** - \( c = 3.00 \times 10^{8} \) meters per second (m/s) - **Electron volt to Joule conversion:** - \( 1 \ \text{eV} = 1.602 \times 10^{-19} \) Joules (J) However, there's a more convenient approach for photon energy calculations in eV when the wavelength is in nanometers: - Use \( hc \) in units of eV·nm: \[ hc \approx 1240 \ \text{eV} \cdot \text{nm} \] This approximation is based on the exact values: \[ hc = (6.626 \times 10^{-34} \ \text{J} \cdot \text{s})(3.00 \times 10^{8} \ \text{m/s}) = 1.986 \times 10^{-25} \ \text{J} \cdot \text{m} \] Then converting Joules to eV and meters to nanometers results in \( hc \approx 1240 \ \text{eV} \cdot \text{nm} \). ### **Step 3: Calculate the Photon Energy** Now, plug in the values: \[ E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1240 \ \text{eV} \cdot \text{nm}}{400 \ \text{nm}} \] \[ E_{\text{photon}} = 3.10 \ \text{eV} \] ### **Answer** **The photon energy is 3.10 electron volts (eV).** --- **Explanation:** By using the simplified form of the equation with \( hc \) in eV·nm, the calculation becomes straightforward when the wavelength is given in nanometers. This method avoids unnecessary conversions and reduces the potential for errors. **Additional Context:** - **Work Function (\( \phi \)):** The work function is the minimum energy required to remove an electron from the surface of a material. In this problem, it's given as \( \phi = 2.93 \ \text{eV} \). While it's not directly needed to calculate the photon energy, it could be relevant for further calculations, such as determining if the photoelectric effect will occur (which it would, since \( E_{\text{photon}} > \phi \)). - **Application:** Knowing the photon energy is crucial in fields like photochemistry, quantum mechanics, and materials science, as it determines how photons interact with matter. |
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| GPT-4o | To calculate the photon energy in electron volts (eV) for light of a given wavelength, we can use the following formula: \[ E = \frac{hc}{\lambda} \] where: - \( E \) is the photon energy, - \( h \) is Planck's constant (\( h = 6.626 \times 10^{-34} \) Js), - \( c \) is the speed of light in a vacuum (\( c = 3.00 \times 10^8 \) m/s), - \( \lambda \) is the wavelength of the light. First, we need to convert the wavelength from nanometers to meters: \[ \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \] Now, we substitute the values into the formula: \[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) (3.00 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \] Let's calculate the numerator: \[ 6.626 \times 10^{-34} \, \text{Js} \times 3.00 \times 10^8 \, \text{m/s} = 1.9878 \times 10^{-25} \, \text{Jm} \] Now, divide by the wavelength in meters: \[ E = \frac{1.9878 \times 10^{-25} \, \text{Jm}}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.9878 \times 10^{-25}}{400 \times 10^{-9}} \] \[ E = \frac{1.9878 \times 10^{-25}}{4 \times 10^{-7}} \] \[ E = \frac{1.9878}{4} \times 10^{-18} \] \[ E = 0.49695 \times 10^{-18} \, \text{J} \] We need to convert this energy from joules to electron volts. The conversion factor is: \[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \] So we divide the energy in joules by this conversion factor: \[ E (\text{eV}) = \frac{0.49695 \times 10^{-18} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \] \[ E (\text{eV}) = \frac{0.49695}{1.602} \times 10^{-18 + 19} \] \[ E (\text{eV}) = \frac{0.49695}{1.602} \times 10^1 \] \[ E (\text{eV}) = 0.3103 \times 10 \] \[ E (\text{eV}) = 3.103 \, \text{eV} \] So, the photon energy for light of wavelength 400 nm is approximately \( 3.10 \, \text{eV} \). |
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| llama3.1-70B-Instruct | To calculate the photon energy in eV, we can use the formula: E = hc/λ where E is the photon energy, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the light. First, we need to convert the wavelength from nanometers to meters: λ = 400 nm = 400 x 10^-9 m = 4 x 10^-7 m Now we can plug in the values: E = (6.626 x 10^-34 J s) x (3 x 10^8 m/s) / (4 x 10^-7 m) E = 4.969 x 10^-19 J To convert this energy from joules to electronvolts, we can divide by the elementary charge (1.602 x 10^-19 C): E = 4.969 x 10^-19 J / (1.602 x 10^-19 C) E = 3.10 eV Therefore, the photon energy is 3.10 eV. |
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| llama3.1-405B-Instruct | To calculate the photon energy in eV, we can use the formula: E = hf where E is the photon energy, h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the light. First, we need to find the frequency of the light. We can do this using the formula: f = c / λ where c is the speed of light (3 x 10^8 m/s) and λ is the wavelength of the light (400 nm = 400 x 10^-9 m). Plugging in the values, we get: f = (3 x 10^8 m/s) / (400 x 10^-9 m) = 7.5 x 10^14 Hz Now, we can plug this value into the first formula to get the photon energy: E = hf = (6.626 x 10^-34 J s) x (7.5 x 10^14 Hz) = 4.969 x 10^-19 J To convert this energy from joules to electronvolts, we can use the conversion factor: 1 eV = 1.602 x 10^-19 J So, E = 4.969 x 10^-19 J / (1.602 x 10^-19 J/eV) = 3.10 eV Therefore, the photon energy is 3.10 eV. |
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| DeepSeek-Coder-V2 | To solve the problem of calculating the photon energy in eV when light of wavelength 400 nm is incident upon lithium (with a work function φ = 2.93 eV), we can follow these steps: 1. **Convert the wavelength to meters**: The wavelength given is 400 nm. Since 1 nanometer (nm) is equal to \(10^{-9}\) meters (m), we convert the wavelength as follows: \[ 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} = 4 \times 10^{-7} \, \text{m} \] 2. **Calculate the photon energy using the formula**: The energy \(E\) of a photon is given by the formula: \[ E = \frac{hc}{\lambda} \] where: - \(h\) is Planck's constant, \(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\) - \(c\) is the speed of light, \(3 \times 10^8 \, \text{m/s}\) - \(\lambda\) is the wavelength of the light, \(4 \times 10^{-7} \, \text{m}\) Substituting the values into the formula: \[ E = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3 \times 10^8 \, \text{m/s})}{4 \times 10^{-7} \, \text{m}} \] 3. **Simplify the expression**: \[ E = \frac{1.9878 \times 10^{-25} \, \text{J} \cdot \text{m}}{4 \times 10^{-7} \, \text{m}} = 4.9695 \times 10^{-19} \, \text{J} \] 4. **Convert the energy from joules to electron volts (eV)**: Since 1 eV is equal to \(1.602 \times 10^{-19} \, \text{J}\), we convert the energy as follows: \[ E = \frac{4.9695 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 3.102 \, \text{eV} \] Therefore, the photon energy in eV is: \[ \boxed{3.102} \] |
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| Qwen2-72B-Instruct | To calculate the photon energy in electronvolts (eV), we use the equation that relates the photon energy (E) to its wavelength (λ) in a vacuum: \[ E = \frac{hc}{\lambda} \] where: - \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} \)), - \( c \) is the speed of light in a vacuum (\( 3.0 \times 10^8 \, \text{m/s} \)), - \( \lambda \) is the wavelength of the light (\( 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)). Substituting the given values: \[ E = \frac{(4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s}) \cdot (3.0 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.2408 \times 10^{-6} \, \text{eV}\cdot\text{m}}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.2408 \times 10^{-6} \, \text{eV}\cdot\text{m}}{4 \times 10^{-7} \, \text{m}} \] \[ E = 3.102 \, \text{eV} \] Thus, the photon energy is approximately 3.102 eV. |
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| baichuan4 | To calculate the photon energy in electronvolts (eV) for light of wavelength 400 nm incident upon lithium, we can use the following formula: \[ E = \frac{hc}{\lambda} \] where: - \( E \) is the photon energy in joules (J), - \( h \) is the Planck's constant (\( 6.626 \times 10^{-34} \) J·s), - \( c \) is the speed of light in vacuum (\( 3.00 \times 10^8 \) m/s), - \( \lambda \) is the wavelength of the light in meters (m). First, convert the wavelength from nm to m: \[ \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \] Now, substitute the values into the formula: \[ E = \frac{6.626 \times 10^{-34} \, \text{J·s} \times 3.00 \times 10^8 \, \text{m/s}}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.9878 \times 10^{-26} \, \text{J}}{1} \] \[ E = 1.9878 \times 10^{-26} \, \text{J} \] To convert this energy from joules to electronvolts, we use the conversion factor that 1 eV is equal to \( 1.602 \times 10^{-19} \) J: \[ E (\text{in eV}) = \frac{E (\text{in J})}{1.602 \times 10^{-19} \, \text{J/eV}} \] \[ E (\text{in eV}) = \frac{1.9878 \times 10^{-26} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \] \[ E (\text{in eV}) = 1.24 \times 10^{-7} \, \text{eV} \] Therefore, the photon energy for light of wavelength 400 nm is approximately \( 1.24 \times 10^{-7} \) eV. |
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| doubao | The energy of a photon is given by E = hf, where h is Planck's constant (6.626 x 10^-34 Js) and f is the frequency. The frequency f is related to the wavelength λ by f = c / λ, where c is the speed of light (3.0 x 10^8 m/s). First, convert the wavelength from nm to m: 400 nm = 400 x 10^-9 m. Then calculate the frequency: f = 3.0 x 10^8 / (400 x 10^-9) = 7.5 x 10^14 Hz. Now calculate the energy in Joules: E = 6.626 x 10^-34 Js x 7.5 x 10^14 Hz = 4.9695 x 10^-19 J. To convert this to eV, divide by 1.602 x 10^-19: E = (4.9695 x 10^-19 J) / (1.602 x 10^-19) = 3.1 eV. |
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| minimax | To calculate the photon energy in electronvolts (eV), we can use the equation that relates the energy of a photon (E) to its wavelength (λ) through Planck's constant (h) and the speed of light (c): \[ E = \frac{hc}{\lambda} \] where: - \( E \) is the energy of the photon in joules (J), - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J·s), - \( c \) is the speed of light in a vacuum (\( 3.00 \times 10^8 \) m/s), - \( \lambda \) is the wavelength of the light in meters (m). First, we need to convert the wavelength from nanometers to meters: \[ \lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} \] Now we can plug in the values: \[ E = \frac{(6.626 \times 10^{-34} \text{ J·s})(3.00 \times 10^8 \text{ m/s})}{400 \times 10^{-9} \text{ m}} \] \[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{400 \times 10^{-9}} \] \[ E = \frac{1.9878 \times 10^{-25}}{400 \times 10^{-9}} \] \[ E = \frac{1.9878 \times 10^{-25}}{4 \times 10^{-7}} \] \[ E = 4.9695 \times 10^{-19} \text{ J} \] To convert the energy from joules to electronvolts, we use the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \): \[ E (\text{in eV}) = \frac{4.9695 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \] \[ E (\text{in eV}) = 3.102 \text{ eV} \] Therefore, the photon energy of light with a wavelength of 400 nm is approximately 3.102 eV. |
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| kimi | To calculate the photon energy in electron volts (eV), you can use the Planck's equation: \[ E = \frac{hc}{\lambda} \] where: - \( E \) is the energy of the photon, - \( h \) is Planck's constant (\(6.62607015 \times 10^{-34}\) Js), - \( c \) is the speed of light in a vacuum (\(2.99792458 \times 10^8\) m/s), - \( \lambda \) is the wavelength of the light. Given that the wavelength \( \lambda \) is 400 nm (nanometers), we first need to convert it to meters: \[ \lambda = 400 \times 10^{-9} \text{ m} \] Now, we can plug in the values: \[ E = \frac{6.62607015 \times 10^{-34} \text{ Js} \times 2.99792458 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} \] \[ E = \frac{6.62607015 \times 2.99792458 \times 10^{-25}}{400} \text{ J} \] \[ E = \frac{1.9865 \times 10^{-19}}{400} \text{ J} \] \[ E = 4.96625 \times 10^{-21} \text{ J} \] To convert the energy from joules to electron volts, we use the conversion factor \(1 \text{ eV} = 1.602176634 \times 10^{-19} \text{ J}\): \[ E_{\text{eV}} = \frac{E}{1.602176634 \times 10^{-19}} \] \[ E_{\text{eV}} = \frac{4.96625 \times 10^{-21}}{1.602176634 \times 10^{-19}} \] \[ E_{\text{eV}} \approx 3.10 \text{ eV} \] So, the photon energy of the light with a wavelength of 400 nm is approximately 3.10 eV. |
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| wenxinyiyan | We have light of wavelength 400 nanometers incident upon lithium, which has a work function (phi) of 2.93 electron volts. Our task is to calculate the energy of the photon in electron volts. To calculate the energy of a photon, we use the formula: E = h × ν where E is the energy of the photon, h is Planck's constant, and ν is the frequency of the light. However, we are given the wavelength of the light, not its frequency. We can calculate the frequency using the formula: ν = c / λ where c is the speed of light and λ is the wavelength of the light. Combining these two formulas, we get: E = h × c / λ Now, let's plug in the values: h (Planck's constant) = 4.135667696 × 10^(-15) eV·s c (speed of light) = 2.99792458 × 10^8 m/s λ (wavelength) = 400 × 10^(-9) m E = (4.135667696 × 10^(-15) eV·s) × (2.99792458 × 10^8 m/s) / (400 × 10^(-9) m) = 3.0996049601375923 eV So, the energy of a photon of light with a wavelength of 400 nm is approximately 3.10 eV. |
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| zhipu | To calculate the photon energy in electron volts (eV) for light of a given wavelength, we can use the following relationship between the energy of a photon (E), the speed of light (c), the wavelength of the light (λ), and Planck's constant (h): \[ E = \frac{hc}{\lambda} \] Where: - \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s), - \( c \) is the speed of light in a vacuum (\( 3 \times 10^8 \) m/s), - \( \lambda \) is the wavelength of the light (400 nm, which is \( 400 \times 10^{-9} \) m). First, let's plug in the values: \[ E = \frac{(4.135667696 \times 10^{-15} \text{ eV·s}) \times (3 \times 10^8 \text{ m/s})}{400 \times 10^{-9} \text{ m}} \] Now, let's do the calculation: \[ E = \frac{(4.135667696 \times 10^{-15}) \times (3 \times 10^8)}{400 \times 10^{-9}} \] \[ E = \frac{(1.2407003088 \times 10^{-6})}{400 \times 10^{-9}} \] \[ E = \frac{1.2407003088 \times 10^{-6}}{4 \times 10^{-7}} \] \[ E = 3.101757572 \text{ eV} \] Rounding to a reasonable number of significant figures (given that the input values have around 3 significant figures), we get: \[ E \approx 3.10 \text{ eV} \] So, the photon energy of light with a wavelength of 400 nm is approximately 3.10 eV. |
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| gpto1 | To calculate the photon energy \( E_{\text{photon}} \) in electron volts (eV) for light of wavelength \( \lambda = 400 \) nanometers (nm), we'll use the fundamental relationship between energy, wavelength, and the constants of physics. ### **Step 1: Understand the Relationship** The energy of a photon is given by the equation: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] Where: - \( E_{\text{photon}} \) is the energy of the photon. - \( h \) is Planck's constant. - \( c \) is the speed of light in a vacuum. - \( \lambda \) is the wavelength of the photon. ### **Step 2: Use Appropriate Units** To ensure consistency, we need to use units that will yield an energy in electron volts (eV). **Important constants:** - **Planck's constant (\( h \)):** - \( h = 6.626 \times 10^{-34} \) Joule\(\cdot\)seconds (J·s) - **Speed of light (\( c \)):** - \( c = 3.00 \times 10^{8} \) meters per second (m/s) - **Electron volt to Joule conversion:** - \( 1 \ \text{eV} = 1.602 \times 10^{-19} \) Joules (J) However, there's a more convenient approach for photon energy calculations in eV when the wavelength is in nanometers: - Use \( hc \) in units of eV·nm: \[ hc \approx 1240 \ \text{eV} \cdot \text{nm} \] This approximation is based on the exact values: \[ hc = (6.626 \times 10^{-34} \ \text{J} \cdot \text{s})(3.00 \times 10^{8} \ \text{m/s}) = 1.986 \times 10^{-25} \ \text{J} \cdot \text{m} \] Then converting Joules to eV and meters to nanometers results in \( hc \approx 1240 \ \text{eV} \cdot \text{nm} \). ### **Step 3: Calculate the Photon Energy** Now, plug in the values: \[ E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1240 \ \text{eV} \cdot \text{nm}}{400 \ \text{nm}} \] \[ E_{\text{photon}} = 3.10 \ \text{eV} \] ### **Answer** **The photon energy is 3.10 electron volts (eV).** --- **Explanation:** By using the simplified form of the equation with \( hc \) in eV·nm, the calculation becomes straightforward when the wavelength is given in nanometers. This method avoids unnecessary conversions and reduces the potential for errors. **Additional Context:** - **Work Function (\( \phi \)):** The work function is the minimum energy required to remove an electron from the surface of a material. In this problem, it's given as \( \phi = 2.93 \ \text{eV} \). While it's not directly needed to calculate the photon energy, it could be relevant for further calculations, such as determining if the photoelectric effect will occur (which it would, since \( E_{\text{photon}} > \phi \)). - **Application:** Knowing the photon energy is crucial in fields like photochemistry, quantum mechanics, and materials science, as it determines how photons interact with matter. |
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| deepseekv3 | To calculate the photon energy in electron volts (eV) for light of a given wavelength, we can use the following formula: \[ E = \frac{hc}{\lambda} \] Where: - \( E \) is the photon energy in joules (J), - \( h \) is Planck's constant (\( h = 6.626 \times 10^{-34} \) J·s), - \( c \) is the speed of light (\( c = 3.00 \times 10^8 \) m/s), - \( \lambda \) is the wavelength of the light in meters (m). Since we want the energy in electron volts (eV), we will convert the energy from joules to eV using the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Given: - Wavelength \( \lambda = 400 \) nm \( = 400 \times 10^{-9} \) m. ### Step 1: Calculate the energy in joules \[ E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \text{ J·s})(3.00 \times 10^8 \text{ m/s})}{400 \times 10^{-9} \text{ m}} \] \[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{400 \times 10^{-9}} \] \[ E = \frac{1.9878 \times 10^{-25}}{400 \times 10^{-9}} \] \[ E = 4.9695 \times 10^{-19} \text{ J} \] ### Step 2: Convert the energy to electron volts \[ E (\text{eV}) = \frac{E (\text{J})}{1.602 \times 10^{-19} \text{ J/eV}} \] \[ E (\text{eV}) = \frac{4.9695 \times 10^{-19}}{1.602 \times 10^{-19}} \] \[ E (\text{eV}) \approx 3.10 \text{ eV} \] ### Final Answer: The photon energy of light with a wavelength of 400 nm is approximately **3.10 eV**. |
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