A mercury atom drops from 8.82 eV to 6.67 eve.
1) What is the energy of the photon emitted by the mercury atom?
2) What is the frequency of the photon?
3)What is the wavelength?
^anon actually meant to say yes
No
To answer your questions, we can make use of the energy-photon relationship given by the equation E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon.
1) To find the energy of the photon emitted by the mercury atom, we can subtract the initial energy from the final energy:
E = 8.82 eV - 6.67 eV = 2.15 eV
Therefore, the energy of the photon emitted by the mercury atom is 2.15 eV.
2) To find the frequency of the photon, we need to convert the energy from electron volts (eV) to joules (J). We can use the conversion factor 1 eV = 1.6022 x 10^-19 J.
E = 2.15 eV * 1.6022 x 10^-19 J/eV = 3.44 x 10^-19 J
Now, we can rearrange the equation E = hf to find the frequency:
f = E / h
where h is Planck's constant, which is approximately 6.626 x 10^-34 J·s.
f = (3.44 x 10^-19 J) / (6.626 x 10^-34 J·s)
f ≈ 5.2 x 10^14 Hz
Therefore, the frequency of the photon emitted by the mercury atom is approximately 5.2 x 10^14 Hz.
3) To find the wavelength of the photon, we can use the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency.
First, we need to convert the frequency to angular frequency (ω) using the equation ω = 2πf:
ω = 2π x (5.2 x 10^14 Hz) = 3.28 x 10^15 rad/s
Now, we can rearrange the equation c = ωλ to find the wavelength:
λ = c / ω
where c is the speed of light, which is approximately 3 x 10^8 m/s.
λ = (3 x 10^8 m/s) / (3.28 x 10^15 rad/s)
λ ≈ 9.15 x 10^-8 m
Therefore, the wavelength of the photon emitted by the mercury atom is approximately 9.15 x 10^-8 m.
To find the energy of the photon emitted by the mercury atom, you can use the formula E = hf, where E is the energy, h is the Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the photon.
1) First, we need to find the difference in energy between the initial and final energy levels of the mercury atom. This can be calculated by subtracting the initial energy (8.82 eV) from the final energy (6.67 eV):
ΔE = 6.67 eV - 8.82 eV = -2.15 eV
Note: Since the energy decreased, the value is negative.
2) To convert the energy from electron volts (eV) to joules (J), we need to use the conversion factor: 1 eV = 1.602 x 10^-19 J. Multiplying the change in energy by the conversion factor gives us the energy in joules:
ΔE = -2.15 eV * (1.602 x 10^-19 J/eV) = -3.443 x 10^-19 J
3) Now, we can calculate the frequency of the photon using the formula E = hf. Rearranging the equation gives f = E/h:
f = (-3.443 x 10^-19 J) / (6.626 x 10^-34 J·s)
f = -5.2 x 10^14 Hz
Note: The negative sign indicates a decrease in energy.
4) Finally, to find the wavelength, we can use the equation c = λf, where c is the speed of light (3 x 10^8 m/s), λ is the wavelength, and f is the frequency of the photon:
λ = c / f
λ = (3 x 10^8 m/s) / (5.2 x 10^14 Hz)
λ ≈ 5.8 x 10^-7 m or 580 nm
So, the answers to your questions are:
1) The energy of the photon emitted by the mercury atom is -3.443 x 10^-19 J.
2) The frequency of the photon is -5.2 x 10^14 Hz.
3) The wavelength of the photon is approximately 5.8 x 10^-7 m or 580 nm.