Module Post Assessment #34B

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joanneyseung22
Posts: 31
Joined: Fri Sep 28, 2018 12:15 am

Module Post Assessment #34B

Postby joanneyseung22 » Thu Oct 18, 2018 12:57 am

Question states that "Molybdenum metal must absorb radiation with a minimum frequency of 1.09 x 1015 s-1 before it can emit an electron from its surface. If molybdenum is irradiated with 194 nm light, what is the maximum possible kinetic energy of the emitted electrons?"
How do you calculate the kinetic energy in this problem? I feel like you should be using 1/2mv^2, but I am not sure how to find the velocity given the 194 nm light.

janeane Kim4G
Posts: 31
Joined: Fri Sep 28, 2018 12:28 am

Re: Module Post Assessment #34B

Postby janeane Kim4G » Thu Oct 18, 2018 3:21 am

Hello!
For this problem, you would use the equations frequency=c/wavelength, E=hv, and E(kinetic)=E(photon)-work function in that order. Use the first equation to and plug in the given 194 nm in correct units to find the frequency of the light ray:
(3*10^8)/(1.94*10^-9 m)= 1.55*10^15 Hz

Then, we can find the energy of the photon by multiplying that number with the plancks constant
1.55*10^-15 x 6.626*10^-34= 1.025*10^-19 J

And using the last equation, subtract this value that we got here and the previous value from part A, which was the minimum power of the photon in order to eject an electron from Molybdenum, which is the work function for this metal.
1.025*10^-18 J- 7.22*-19 J= 3.03*10^-19 J

This was how to solve it but I got about 2 decimal places off, so you might have to leave some constants in more sig figs to get the exact module answer but I’m pretty sure there was something similar to mine and it was right (?) hope this helps !

Stevin1H
Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

Re: Module Post Assessment #34B

Postby Stevin1H » Thu Oct 18, 2018 10:30 am

In order to solve this problem, you would also need the answer from part A. Part A asks for the the amount of energy to emit an electron with a frequency of 1.09E15 1/s. Given this frequency, use the equation c=(lambda)(frequency) and solve for lambda. lambda=(speed of light)/(frequency) = 2.75 E-7m. Given the wavelength, plug this value into the equation E=(Planck's constant)(speed of light)/(wavelength) to find the energy that is required to eject the electron. Your value should equal 7.22E-19.

For part B, find the energy of the photon given the wavelength of 194nm. To do so, use the equation E=(Planck's constant)(speed of light)/(wavelength). Your answer should be 1.02E-18J. Given the amount of energy from the photon, subtract this value from the threshold energy (the amount of energy to eject an electron) to find the amount of kinetic energy, which is equal to 3.05E-19J.

E(photon) - Threshold energy = Kinetic energy.


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