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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: schnappy on November 07, 2010, 10:06:24 pm

Title: Finding energy of electron, given lambda
Post by: schnappy on November 07, 2010, 10:06:24 pm
Hey all,
I've done this question before but lost my formula sheet i had at the time.

All you're told is that de Broglie lambda = 0.2 nm.

Then find momentum, and then the energy in eV of the electron.

I've found the momentum to be 7.32E-24.

But I'm having trouble getting the correct energy.

Edit: could someone please upload the answers to 2010 Lisachem unit4 physics exam? Thanks.
Title: Re: Finding energy of electron, given lambda
Post by: Linkage1992 on November 07, 2010, 10:09:32 pm
Use hc / lambda.

see below!  :)
Title: Re: Finding energy of electron, given lambda
Post by: schnappy on November 07, 2010, 10:10:39 pm
Quote from: Linkage1992 on November 07, 2010, 10:09:32 pm
Use hc / lambda.

No. That's fundamentally wrong since electrons do not go at the speed of light...

Every method I've tried i get 2.34E20. It sounds wrong.... but is it?
Title: Re: Finding energy of electron, given lambda
Post by: 3Xamz on November 07, 2010, 10:13:01 pm
Could you upload the question?
I'd have a crack at it
Title: Re: Finding energy of electron, given lambda
Post by: Linkage1992 on November 07, 2010, 10:13:34 pm
woops sorry, wasn't thinking straight.

What you have to do work out the velocity of the electron. You have the mass and the momentum so you can use P = mv.
Then use E = .5mv^2 to get the kinetic energy which will be in joules.
Then convert this to eVs.
Title: Re: Finding energy of electron, given lambda
Post by: schnappy on November 07, 2010, 10:17:25 pm
Arr I see what I did wrong, thanks. Work in joules, until the end convert to eV. It's been way to long since I last did this!
Title: Re: Finding energy of electron, given lambda
Post by: schnappy on November 07, 2010, 10:23:42 pm
Actually, I'm still getting it wrong. Can someone PLEASE do it? :(
Title: Re: Finding energy of electron, given lambda
Post by: Whatlol on November 07, 2010, 10:29:56 pm
lambda = h/p
therefore p = 6.63x10^-34 / 0.2x10^-9
             p=3.315x10^-24 kgms^-1
             v= p/m
             v = 3.315x10^-24 /9.1x10^-31
             v = 3642857ms^-1
             Ek =1/2mv^2
             Ek = 1/2 x 9.1x10^-31 x 3652857 ^2
             Ek = 6.038x10^-18 J
            and that is equivilent to 37.7 eV
Title: Re: Finding energy of electron, given lambda
Post by: schnappy on November 07, 2010, 10:36:40 pm
Thanks. I asked someone for the answers to the question... I used a result to a previous question. My answer was correct there, but misread a 3 as a 7.
Title: Re: Finding energy of electron, given lambda
Post by: Whatlol on November 07, 2010, 10:40:40 pm
Quote from: schnappy on November 07, 2010, 10:36:40 pm
Thanks. I asked someone for the answers to the question... I used a result to a previous question. My answer was correct there, but misread a 3 as a 7.

heres the exam with answers at the back
Title: Re: Finding energy of electron, given lambda
Post by: shokstar on November 08, 2010, 04:02:04 pm
Quote from: Whatlol on November 07, 2010, 10:29:56 pm
lambda = h/p
therefore p = 6.63x10^-34 / 0.2x10^-9
             p=3.315x10^-24 kgms^-1
             v= p/m
             v = 3.315x10^-24 /9.1x10^-31
             v = 3642857ms^-1
             Ek =1/2mv^2
             Ek = 1/2 x 9.1x10^-31 x 3652857 ^2
             Ek = 6.038x10^-18 J
            and that is equivilent to 37.7 eV

Or just  Ek=(p2)/2m, and from Ek to p=root(2mXEk)