problem 1.69

Moderators: Chem_Mod, Chem_Admin

Madeline_Foo_3J
Posts: 26
Joined: Wed Sep 21, 2016 3:00 pm

problem 1.69

Postby Madeline_Foo_3J » Sat Oct 08, 2016 9:17 pm

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 (694nm) and a low intensity violet GaN laser (405nm), but (a) which laser they should use and (b) the kinetic energy of the electrons emitted. The work function of lithium is 2.93eV.

My questions is how are we supposed to set it up? What are we supposed to do with the eV units?

Hao 1I
Posts: 26
Joined: Wed Sep 21, 2016 3:00 pm
Been upvoted: 1 time

Re: problem 1.69

Postby Hao 1I » Sun Oct 09, 2016 9:27 pm

First, you need to convert 2.93 eV to joules because work function should be in joules. 1eV is 1.602x10^-19 J.
And to set this up, you need to compare which laser provides the most energy.


Return to “Photoelectric Effect”

Who is online

Users browsing this forum: No registered users and 2 guests