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ChatGPT
Jun 3
To calculate the wavelength of the light emitted when an atom moves from the ground level to the highest energy level, we can use the formula:
ΔE = hc/λ
Where:
ΔE = change in energy level (in joules)
h = Planck's constant (6.626 x 10^-34 J s)
c = speed of light (3.00 x 10^8 m/s)
λ = wavelength of light emitted (in meters)
First, we need to calculate the change in energy level:
ΔE = 3.00 eV - 1.00 eV
ΔE = 2.00 eV
Next, we need to convert the energy change from electron volts (eV) to joules:
1 eV = 1.602 x 10^-19 J
2.00 eV = 2.00 x 1.602 x 10^-19 J
ΔE = 3.204 x 10^-19 J
Now, we can plug in the values into the formula to calculate the wavelength:
3.204 x 10^-19 J = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s) / λ
λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s) / 3.204 x 10^-19 J
λ = 6.24 x 10^-7 meters
Therefore, the wavelength of the light emitted when the atom moves from the ground level to the highest energy level is 624 nm.
ΔE = hc/λ
Where:
ΔE = change in energy level (in joules)
h = Planck's constant (6.626 x 10^-34 J s)
c = speed of light (3.00 x 10^8 m/s)
λ = wavelength of light emitted (in meters)
First, we need to calculate the change in energy level:
ΔE = 3.00 eV - 1.00 eV
ΔE = 2.00 eV
Next, we need to convert the energy change from electron volts (eV) to joules:
1 eV = 1.602 x 10^-19 J
2.00 eV = 2.00 x 1.602 x 10^-19 J
ΔE = 3.204 x 10^-19 J
Now, we can plug in the values into the formula to calculate the wavelength:
3.204 x 10^-19 J = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s) / λ
λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s) / 3.204 x 10^-19 J
λ = 6.24 x 10^-7 meters
Therefore, the wavelength of the light emitted when the atom moves from the ground level to the highest energy level is 624 nm.
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Google Gemini AI
Jun 3
The energy of a photon is given by:
$$E = hf = \frac{hc}{\lambda}$$
where:
- $E$ is the energy of the photon (in joules)
- $h$ is Planck's constant
$$h=6.626\times10^{-34}Js$$
- $f$ is the frequency of the photon (in hertz)
- $c$ is the speed of light
$$c=3\times10^{8}m/s$$
- $\lambda$ is the wavelength of the photon (in meters)
The change in energy of an atom when it moves from one energy level to another is given by:
$$\Delta E = E_f - E_i$$
where:
- $\Delta E$ is the change in energy (in joules)
- $E_f$ is the energy of the final energy level (in joules)
- $E_i$ is the energy of the initial energy level (in joules)
In this case, the atom moves from the ground level (1.00 eV) to the highest energy level (3.00 eV), so the change in energy is:
$$\Delta E = 3.00 eV - 1.00 eV = 2.00 eV$$
Converting the energy to joules:
$$\Delta E = 2.00 eV \times 1.602 \times 10^{-19} J/eV = 3.204 \times 10^{-19} J$$
The wavelength of the light emitted can now be calculated:
$$\lambda = \frac{hc}{\Delta E}$$
Substituting the values:
$$\lambda = \frac{(6.626 \times 10^{-34} J s)(3 \times 10^8 m/s)}{3.204 \times 10^{-19} J}$$
$$\lambda =632.8 \times 10^{-9} m = 632.8 nm$$
Therefore, the wavelength of the light emitted is 632.8 nm.
$$E = hf = \frac{hc}{\lambda}$$
where:
- $E$ is the energy of the photon (in joules)
- $h$ is Planck's constant
$$h=6.626\times10^{-34}Js$$
- $f$ is the frequency of the photon (in hertz)
- $c$ is the speed of light
$$c=3\times10^{8}m/s$$
- $\lambda$ is the wavelength of the photon (in meters)
The change in energy of an atom when it moves from one energy level to another is given by:
$$\Delta E = E_f - E_i$$
where:
- $\Delta E$ is the change in energy (in joules)
- $E_f$ is the energy of the final energy level (in joules)
- $E_i$ is the energy of the initial energy level (in joules)
In this case, the atom moves from the ground level (1.00 eV) to the highest energy level (3.00 eV), so the change in energy is:
$$\Delta E = 3.00 eV - 1.00 eV = 2.00 eV$$
Converting the energy to joules:
$$\Delta E = 2.00 eV \times 1.602 \times 10^{-19} J/eV = 3.204 \times 10^{-19} J$$
The wavelength of the light emitted can now be calculated:
$$\lambda = \frac{hc}{\Delta E}$$
Substituting the values:
$$\lambda = \frac{(6.626 \times 10^{-34} J s)(3 \times 10^8 m/s)}{3.204 \times 10^{-19} J}$$
$$\lambda =632.8 \times 10^{-9} m = 632.8 nm$$
Therefore, the wavelength of the light emitted is 632.8 nm.
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