Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. What is the energy of a photon of this light, in eV?
E = hc/wavelength
wavelength must be converted to m.
E is in joules. Convert to eV.
To calculate the energy of a photon, you can use the equation:
E = hc / λ
where E is the energy of the photon, h is Planck's constant (approximately 6.626 x 10^-34 J.s), c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s), and λ is the wavelength of the light.
First, let's convert the given wavelength from nanometers (nm) to meters (m):
58.4 nm = 58.4 x 10^-9 m
Now, we can substitute the values into the equation:
E = (6.626 x 10^-34 J.s * 3.00 x 10^8 m/s) / (58.4 x 10^-9 m)
Simplifying the expression:
E = (1.99 x 10^-25 J.m) / (58.4 x 10^-9 m)
E = 3.41 x 10^-17 J
To convert the energy from joules to electron volts (eV), we can use the conversion factor:
1 eV = 1.6 x 10^-19 J
Dividing the energy by this conversion factor, we get:
E = (3.41 x 10^-17 J) / (1.6 x 10^-19 J/eV)
E ≈ 213 eV
Therefore, the energy of a photon with a wavelength of 58.4 nm is approximately 213 eV.