1B15


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Amy Luu 2G
Posts: 92
Joined: Wed Sep 18, 2019 12:19 am

1B15

Postby Amy Luu 2G » Wed Oct 16, 2019 11:08 pm

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 103 km/s
(a) What is the wavelength of the ejected electron?

I initially solved this problem by finding the kinetic energy using the equation Ek= (1/2)mv2 and got 5.9 x 10-18 J.
I then found the wavelength using the equation to get 3.4 X 10-8 which is not the correct answer. I know to find the wavelength you use the equation [tex]\lambda=h/mv to get 2.0 x 10^10. How come the first method I used is incorrect? I was expecting to get the same answer even though i approached the problem differently. Can someone explain why what I did was incorrect?

nicolely2F
Posts: 143
Joined: Sat Sep 14, 2019 12:17 am

Re: 1B15

Postby nicolely2F » Thu Oct 17, 2019 12:40 am

In the equation , E is the total energy of the photon, so plugging in the value of the ejected electron's kinetic energy was incorrect. To find the wavelength you can simply use De Broglie's Equation, , where m is the mass of the electron and V is the velocity (in meters!). Plug in the values and you should get the correct answer.

VPatankar_2L
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Joined: Thu Jul 25, 2019 12:17 am
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Re: 1B15

Postby VPatankar_2L » Thu Oct 17, 2019 12:04 pm

Yes, to get the wavelength of the electron you can't use c = 3 * 10^8 because it is the speed of light, thus the speed of a photon. The question has specified the velocity of the electron, so you have to use this velocity in De Broglie's equation.

Michelle Le 1J
Posts: 50
Joined: Fri Aug 09, 2019 12:16 am

Re: 1B15

Postby Michelle Le 1J » Thu Oct 17, 2019 1:19 pm

Kinetic energy(Ek) is different than the E in lambda=hc/E, in which E is the energy of the photon. If you remember the equation we learned while learning the photoelectic effect, it was E(photon) -work function= Ek. Because of the work function, E(photon) is not the same thing as Ek, so that's why your calculation wasn't correct when you first did it.


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