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Topic: de Broglie and electron energy  (Read 8448 times)

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lmvent

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de Broglie and electron energy
« on: October 19, 2005, 08:36:34 PM »
What is the de Broglie wavelength of an oxygen molecule, O2, traveling at 285 m/s? Give answer in pm. Do not enter unit.

I used the equation lambda = (h/(mv)) using 0.032 kg for m, 285 m/s for v and 6.63 x 10^-34 for h.  I get lambda to be 7.27 x 10^-35 m and then divide by 1 x 10^-9 to convert to picometers.  Anyone know what I am doing wrong?

Light with a wavelength of 411 nm fell on a strontium surface, and electrons were ejected. If the speed of an ejected electron is 3.26×105 m/s, what energy was expended in removing the electron from the metal? Express the answer in joules (per electron). Do not enter unit.

I tried to mix a few of the equations given: E = hf, c = lambda*f, and lambda = (h/(mv)) but all of the variables of the last equation are known and the equation is not true.  I used lambda = 411 x 10^-9, h = 6.63 x 10^-34, m = 9.1 x 10^-31 and v = 3.26 x 10^5.  Anyone know what I am doing wrong?

Mitch

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Re:de Broglie and electron energy
« Reply #1 on: October 19, 2005, 08:50:31 PM »
pico is 10^-12m not 10^-9m
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lmvent

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Re:de Broglie and electron energy
« Reply #2 on: October 19, 2005, 09:07:45 PM »
I lied I did use 1 x 10^-12, but it still doesn't work.

Mitch

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Re:de Broglie and electron energy
« Reply #3 on: October 19, 2005, 09:19:04 PM »
No one likes liars. You said you divided by 1 x 10^-12 above, but you actually have to multiply.
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lmvent

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Re:de Broglie and electron energy
« Reply #4 on: October 19, 2005, 09:51:41 PM »
It is still wrong, but why do you multiply?

7.27E-35 m x (1 pm/(10^-12 m )) = pm right??

Mitch

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Re:de Broglie and electron energy
« Reply #5 on: October 19, 2005, 10:20:44 PM »
But the meters are in the denominator not the numerator, so everything is inversed.
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lmvent

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Re:de Broglie and electron energy
« Reply #6 on: October 19, 2005, 10:22:26 PM »
If lambda = (h/(mv)) then lambda is in the numerator isn't it?

Mitch

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Re:de Broglie and electron energy
« Reply #7 on: October 19, 2005, 10:24:29 PM »
Nevermind, I thought you were converting the velocity, which is in the denominator on the other side..
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Mitch

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Re:de Broglie and electron energy
« Reply #8 on: October 19, 2005, 10:25:14 PM »
By the way your mass for a single molecule of O2 is way too large.
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lmvent

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Re:de Broglie and electron energy
« Reply #9 on: October 19, 2005, 10:29:21 PM »
Isn't it 32 grams, so 0.032 kg??

lmvent

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Re:de Broglie and electron energy
« Reply #10 on: October 19, 2005, 10:31:36 PM »
Nevermind I get that. Thanks. Can I get a hint for the other problem??

Mitch

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Re:de Broglie and electron energy
« Reply #11 on: October 19, 2005, 10:46:32 PM »
Its just the energy of 411nm light minus the energy of a 3.26×10^5 m/s electron.
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lmvent

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Re:de Broglie and electron energy
« Reply #12 on: October 19, 2005, 10:53:08 PM »
How do I find the energy of the electron? Is it (1/2)*m*v^2 ?? And I was wrong I am still getting the first problem wrong using (0.032/(6.02 x 10^23)) for the mass.

Mitch

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Re:de Broglie and electron energy
« Reply #13 on: October 19, 2005, 11:24:04 PM »
.5mv^2 sounds right.
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