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Offline xstrae

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Atomic Physics doubt
« on: August 06, 2006, 10:14:51 AM »
I have a doubt regarding certain problems which require you to find the accelerating potential that must be imparted to a proton beam to give it to an effective wavelength of "some value" eg 0.005nm.

I have been taught two formulas from different sources to solve this problem, each giving two solutions. Can anyone please give tell me which of these is correct?

1)  x = h / (root of 2meV) where 'h' is planck's constant, 'm' is the mass of the particle, 'e' is the    charge and 'V' is the potential difference.

2)  x = root of (150/V) amstrong units ,where 'x' is the wavelength and 'V' is the potential difference.

Here is the derivation for the above two formulas:

1)  E = P2/2m
     P = root of (2mE)
     x = h/p
     x = h/root of (2mE) 
     x = h/root of (2meV)  [by the way is 'V' here in electrostatic units or in couloumbs. My teacher said voltage in electrostatic unit * 1/3 * 10-2 = V in couloumbs.Is that correct? ???]


2) E=Ve
Therefore, 1/2mv2 = Ve
v2 = 2Ve/m
v = root of (2Ve/m)
mv = m * root (2Ve/m)       [multiplying by 'm' on both sides]
mv = root of (2Vem)
x = h/mv = h/ root of (2Vem)
x = root of (h2/2Vem)

Substituting values of h, m and e,
x = root of (150/V) * 10-10m
x = root of (150/V) Amstrong

Can someone please guide me? ???
« Last Edit: August 06, 2006, 11:52:56 AM by konichiwa2x »

Offline Borek

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Re: Atomic Physics doubt
« Reply #1 on: August 06, 2006, 10:36:11 AM »
I am too lazy to check derivations (it is slow Sunday afternoon after all) but both formulas look similar to me, if you compare them and cancel out V-1/2 looks like

h/(2me)1/2 = 1501/2

Just check units and values to see if it is true.
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Offline xstrae

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Re: Atomic Physics doubt
« Reply #2 on: August 08, 2006, 10:37:31 AM »
then why is it that i get different answers using each one and both turn out to be wrong??
The question was:

calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005nm.      (answer: 32.85V)

using the first formula, i got 7 * 1013V.
and with the second, i got 5 * 1046V.

please help.


Offline Borek

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Re: Atomic Physics doubt
« Reply #3 on: August 08, 2006, 11:24:19 AM »
Looking through both derivations you can find that

1)  x = h/root of (2meV)

2) x = h/ root of (2Vem)

they are equivalent, as I suspected. Check your math.

Although I must admit I am unable to get answer you have listed, looks to me like it should be 3.28V, not 32.8.
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Offline xiankai

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Re: Atomic Physics doubt
« Reply #4 on: August 08, 2006, 11:41:32 AM »
difference between two derivations is trivial; there is just one additional step of including h into the square root formula in the second method.

whether units of V is in electrostatic units (c.g.s. or centimeter, gram, second set of units) or volts (SI set of units, not the coulomb set of units  :))  is not important; using either means you have to use the same units for e too.

i got the correct answer, using my own values from my school's data booklet:

h = 6.63 x 10-34 J s-1 (the Planck constant)

mp = 1.67 x 10-27 kg (rest mass of proton, 11H)

e = -1.60 x 10-19 C (electronic charge)

note that i used coulombs since they asked for volts. usage of appropriate units is important
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Offline Borek

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Re: Atomic Physics doubt
« Reply #5 on: August 08, 2006, 12:40:14 PM »
V = h2/(2me?2)

OK, mass of proton is 1.673*10-27, not 1.673*10-26, that's why I was ten times off.
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Offline xstrae

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Re: Atomic Physics doubt
« Reply #6 on: August 09, 2006, 03:34:25 AM »
I got it now, thanks a lot guys!  ;D

by the way, I was getting the wrong answer using the second formula because when I earlier substitued the values during the derivation, I replaced 'm' with 9.1 * 10-31. So that one would be applicable only for problems involving an electron being accelerated.

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