## 1.31

$c=\lambda v$

RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

### 1.31

How do you find the kinetic energy of the electrons emitted?

"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."

Ethan Lam 4A
Posts: 69
Joined: Thu Jul 11, 2019 12:17 am

### Re: 1.31

First you need to convert the eV into Joules. Then you use that value in Joules in the formula Ek=hc/(wavelenth).

Noah Canio 3C
Posts: 55
Joined: Wed Sep 18, 2019 12:20 am

### Re: 1.31

Recall this equation:

$E_{photon} - Work function = E_{electron}$

You're supposed to determine which laser to use. Take the appropriate laser's wavelength and use u = C/ lamda to solve for the energy of the photon.

Subtract the energy by the work function to find the energy emitted by the electron.

GFolk_1D
Posts: 101
Joined: Fri Aug 09, 2019 12:15 am

### Re: 1.31

Posts: 49
Joined: Wed Sep 18, 2019 12:20 am

### Re: 1.31

Work = threshold energy.
First, you need to convert the work from ev to joules:

2.93eV * (1.602*10^-19J/eV) = 4.69*10^-19J

Second, convert each wavelength of given into energy using E=hc/(wavelength):

2.86*10^-19 (red ruby) and 4.91*10^-19 J (GaN)

Third, compare the energies you calculated to the threshold energy. The wavelength of the choice light should have had to have enough energy to surpass the threshold energy:

4.91*10^-19 J > 4.69*10^-19 J, GaN is the choice light/laser.