1B.15 c)


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Labiba Sardar 2A
Posts: 103
Joined: Sat Jul 20, 2019 12:15 am

1B.15 c)

Postby Labiba Sardar 2A » Sun Oct 13, 2019 12:02 pm

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 10^km s^-1. (a) What is the wavelength of the ejected electron? ... (c) What is the wavelength of the radiation that caused photoejection of the electron?

What's the difference between the wavelengths in parts (a) and (c), conceptually and how you solve for them?

Dan M -3E
Posts: 101
Joined: Wed Sep 18, 2019 12:19 am

Re: 1B.15 c)

Postby Dan M -3E » Sun Oct 13, 2019 3:31 pm

The first one (a) is asking about the wavelength of the electron, and I think that (c) is asking about the wavelength of the photon that caused the electron to be ejected

Isabel Day 1D
Posts: 48
Joined: Fri Aug 09, 2019 12:15 am

Re: 1B.15 c)

Postby Isabel Day 1D » Sun Oct 13, 2019 3:37 pm

The ejected electron is the electron that is released from the metallic surface. The UV radiation is the source of light that caused the ejection of the electron. In order to solve for the wavelength of the emitted electron, find the kinetic energy (KE=1/2mv^2) and use this energy in the equation E=hc/lamda (where h=plank's constant, c= speed of light, and lamda=wavelength). Solve for lamda (or wavelength). To find the wavelength off the incoming UV light, use the equation E(incoming photon)= (work function of the metal: should be given)+(kinetic energy of the outgoing electron: 1/2mv^2). The question should have given you a work function, and we know that since the ejected electron has a positive velocity, the energy of the incoming radiation exceeded the threshold energy. Once you solve for the energy of the incoming electron, you can use the equation E=hc/lamda (where h=plank's constant, c= speed of light, and lamda=wavelength) again to find the wavelength of UV light.

EricZhao3G
Posts: 51
Joined: Tue Oct 08, 2019 12:16 am

Re: 1B.15 c)

Postby EricZhao3G » Mon Oct 14, 2019 2:51 pm

Part a asks for the wavelength of the ejected electron which can be found using the kinetic energy of the ejected electron and E = hv and c = vλ. Part c asks for the wavelength of the proton that caused the electron to be ejected which would be solved in a similar fashion after finding the energy of the proton using E(proton) - E(threshold) = E (kinetic).


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