## 1.69 [ENDORSED]

$E=hv$

peytonruiz 1H
Posts: 20
Joined: Fri Sep 29, 2017 7:03 am
Been upvoted: 1 time

### 1.69

Need help on this problem from the textbook:

1.69) 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.

Nancy Le - 1F
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am

### Re: 1.69  [ENDORSED]

a) An electron can only be ejected if a photon has the minimum energy equal to that of the work function for that metal strikes it.
1. E(work) = (2.93 eV)(1.602 18 x10^(19)JeV^(-1)) = 4.69 x 10^(-19)J
2. Calculate the energy for the red and violet laser using the equation E=hcλ ^(-1).
3. The laser with energy that's greater than or equal to 4.69 x 10^(-19)J will provide enough energy to eject the electron.

b) E(k) can be calculated by subtracting the work function from the energy supplied by the laser:
E(k)= hcλ^(-1) - E(work)

Hope this helps!