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Postby RRahimtoola1I » Sun Oct 27, 2019 11:31 am

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
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Re: 1.31

Postby Ethan Lam 4A » Sun Oct 27, 2019 2:54 pm

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
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Re: 1.31

Postby Noah Canio 3C » Sun Oct 27, 2019 3:30 pm

Recall this equation:

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.

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Re: 1.31

Postby GFolk_1D » Mon Oct 28, 2019 9:29 am

Could someone please explain part a as well please? I am confused about this question.

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Re: 1.31

Postby Varadhan_3G » Mon Oct 28, 2019 5:11 pm

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.

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