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Postby Melissa_Aguirre1J » Sat May 05, 2018 11:58 pm

1.69 In a recent suspense fi lm, 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.

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Re: 1.69

Postby annie_finneran_1K » Sun May 06, 2018 7:40 am

what part are you confused on for this problem? A good hint is to use the equations Ephoton=work function + KE of electron

and E=hv

Tarek Abushamma
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Re: 1.69

Postby Tarek Abushamma » Sun May 06, 2018 11:08 am

In order to solve this problem first convert the 2.93 eV to J. Then, convert both given wavelengths into J using E=hv to see which one corresponds to an energy greater than the 2.93 eV. After you decide which laser has the higher E, subtract the work function from this value in order to find the energy kinetic.

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