Homework 1.69 [ENDORSED]

[Jimmy Zhou 3D]

Question is as follows:

In a recent suspense fi lm, two secret agents must penetrate
a criminal’s stronghold monitored by a lithium photomultiplier
cell that is continually bathed in light from a laser. If the beam
of light is broken, an alarm sounds. The agents want to use a
handheld laser to illuminate the cell while they pass in front
of it. They have two lasers, a high-intensity red ruby laser
(694 nm) and a low-intensity violet GaN laser (405 nm), but
they disagree on which one would be better. Determine
(a) which laser they should use and (b) the kinetic energy of
the electrons emitted. The work function of lithium is 2.93 eV.

I'm not sure how exactly to approach this problem as I'm not sure what qualities would deem a laser as "better"

[Chem_Mod]

[Maggie Bui 1H]

A general approach I typically use is writing down given information and thinking about what I know about the quantities given.

The problem gives us the wavelengths of light and mentions a lithium photomultiplier is used; I singled out these tidbits of the problem because they deal directly with the problem at hand: getting into the stronghold without breaking the beam of light. A photomultiplier detects light by absorbing photons and then emitting electrons, which calls to mind the photoelectric effect. To connect the photomultiplier and wavelength, think about absorption spectra: all elements have unique absorption spectra and and only absorb light of certain wavelengths. The fact that photomultiplier uses lithium is particularly telling. If the secret agents don't want to break the beam of light, they need to use light of a wavelength that will be absorbed by the lithium photomultiplier.

[Jimmy Zhou 3D]

Thanks for the Responses,

Maggie, your explanation makes sense however, the absorption spectra of the lithium photomultiplier is not known. I assumed from Chem Mod's work that as long as the energy emitted by the laser is higher than that of the work function, it should be fine.

Jimmy Z.

[Jimmy Zhou 3D]

Hopefully we will also be able to know how many joules per eV to make the conversion