## Problem 1.31

$c=\lambda v$

josmit_1D
Posts: 92
Joined: Fri Aug 09, 2019 12:17 am

### Problem 1.31

Can someone explain how to solve the following problem: 1.31 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.?

Ryan 1K
Posts: 75
Joined: Fri Aug 09, 2019 12:15 am

### Re: Problem 1.31

To solve this, we must calculate the energy of the photons of each laser.

First, we start with the high-intensity red ruby laser and calculate the frequency.

$c = \lambda v$
$3.00 \times 10^{8} m/s = (694 \times 10^{-9} m)(v)$
$v = 4.32 \times 10^{14} Hz$

Then, we substitute the frequency to calculate E(photon).

$E = hv$
$E = (6.626 \times 10^{-34} Js)(4.32 \times 10^{14} Hz)$
$E = 2.86 \times 10^{-19} J$

In order for the laser to eject electrons, the energy of each photon must be equal to or greater than the work function.

$\phi = 2.93 \times 1.602 \times 10^{-19}$
$\phi = 4.69 \times 10^{-19} J$
$2.86 \times 10^{-19} J < 4.69 \times 10^{-19} J$
$E(photon) < \phi$

This allows us to conclude for part a that they should not use the high-intensity red ruby laser.

To complete part b, we repeat the above steps, this time using the wavelength of the GaN laser (405 nm). We then subtract the work function from the E(photon) to calculate the kinetic energy of the electrons emitted.

$(4.91 \times 10^{-19} J) - (4.69 \times 10^{-19} J) = Ek$
$Ek = 2.14 \times 10^{-20} J$

The answer in the textbook is $2.10 \times 10^{-20} J$, but this is likely due to differences in rounding and conversions.