Wavelength and Kinetic Energy

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Wavelength and Kinetic Energy

Postby KatarinaReid_3H » Mon Oct 05, 2020 4:32 pm

For reference, this question is from Focus 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.

I was confused on both Part A and B. I got that the violet GaN laser does have a higher energy than the lithium so therefore the laser could fill in for the beam of light where the red laser had too low of an energy. My number was 4.91x10^-19J for the GaN laser. According to the textbook answer, the GaN laser had 2.10x10^-20J, therefore wouldn't that be too low of energy also?

Benjamin Chen 1H
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Re: Wavelength and Kinetic Energy

Postby Benjamin Chen 1H » Mon Oct 05, 2020 4:42 pm

I just wanted to point out, it says the electron emitted will have 2.10E-20 J in Kinetic Energy. I also got 4.9 E-19 J for the Energy of GaN.

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Re: Wavelength and Kinetic Energy

Postby AndrewNguyen_2H » Mon Oct 05, 2020 6:47 pm

The GaN laser has 4.9 E-19 J so it is powerful enough to displace electrons. You got part A correct, but when you looked at the answer in the text you looked at the answer for part B. The kinetic energy of the displaced electron will be 2.10E-20 J.
Part B is solved when you subtract the work-energy (2.93 eV) from the energy of the GaN photons (4.9 E-19 J).

Kamille Kibria 2A
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Re: Wavelength and Kinetic Energy

Postby Kamille Kibria 2A » Mon Oct 12, 2020 11:15 am

You got the first part of the question right because the GaN laser had the higher energy of 4.91x10^-19 J. But i think you looked at the answer for part B because the KE of the electrons ejected is actually 2.10x10^-20 J. you can get to that answer by subtracting the work function of lithium from the Ephoton. don't forget to convert from electron volts to joules for the work function though! 1eV=1.6x10^-19 J so for the work function you get 4.69x10^-19 J. after you subtract the work function in J from Ephoton, you end up with the KE of the electrons which is the answer from the book you saw, 2.10x10^-20 J. hope that helps :)

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