Textbook question 1.31

Moderators: Chem_Mod, Chem_Admin

Paige Lee 1A
Posts: 106
Joined: Sat Sep 07, 2019 12:16 am

Textbook question 1.31

Postby Paige Lee 1A » Sat Oct 26, 2019 9:57 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 hand- held 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.

Could someone please explain why the violet laser is the correct answer for part a?

Posts: 66
Joined: Sat Aug 24, 2019 12:15 am

Re: Textbook question 1.31

Postby vpena_1I » Sat Oct 26, 2019 10:15 pm

Using E=hv/入, you find that the energy of the red ruby laser is 2.86x10^-19 J, and the energy of the GaN laser is 4.90x10^-19 J. The work function can be converted to J by using 1 eV=1.602x10^-19 J.
2.93 eV(1.602x10^-19)= 4.69x10^-19J.
This means a photon would need at least 4.69x10^-19 J of energy to eject an electron. The laser that satisfies this is the GaN laser.

Return to “Properties of Electrons”

Who is online

Users browsing this forum: No registered users and 1 guest