We’re being asked to calculate the energy of a photon (eV) with a wavelength of 58.4 nm.
Recall that the energy of a photon (E) is given by:
h = Planck’s constant (6.626 × 10–34 J • s)
v = frequency (in s–1)
Also, recall that the frequency (v) and wavelength (λ) are related:
c = speed of light (3.0 × 108 m/s)
One way to measure ionization energies is photoelectron spectroscopy (PES), a technique based on the photoelectric effect. In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). 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?
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the The Energy of Light concept. You can view video lessons to learn The Energy of Light. Or if you need more The Energy of Light practice, you can also practice The Energy of Light practice problems.