## Question 1.69 Part B (Sixth Edition)

$E=hv$

Steve Magana 2I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Question 1.69 Part B (Sixth Edition)

Question: In a recent suspense film, 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.

Can someone help me understand how to do Part B? I keep getting a different answer, thanks!

2c_britneyly
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

### Re: Question 1.69 Part B (Sixth Edition)

From part (a), you know that the agents should use the low-intensity GaN laser.

For part (b), you want to use the equation Ephoton-work function=Ek to find the kinetic energy emitted. Since you know the wavelength, you can use the equation E=hv to find energy per photon. You are given the work function, so you would subtract the given work function value from Ephoton.