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Quick De Broglie Wavelength Question

I'm doing a question which says to calculate the de broglie wavelength of a proton at speed v.
I have to use the formula -

? = h / (2MeV)^0.5 e - charge of an electron, M - mass, V - accelerating voltage

That's quite badly typed out, i'm afraid.
Should i be using the mass of a proton or electron?

Thank youuu
-Afsaneh-

Reply 1

jaroc

Original post by Churchgirl

Mass of a proton, definitely. The de broglie wavelength of a proton is Planck's constant over momentum of the proton, and momentum of a proton has nothing to do with the mass of an electron :smile:

That's right that

e

in your formula is the charge of an electron, but it's the charge of a proton as well.

(edited 13 years ago)

Reply 2

Churchgirl

Original post by jaroc

Mass of a proton, definitely. The de broglie wavelength of a proton is Planck's constant over momentum of the proton, and momentum of a proton has nothing to do with the mass of an electron :smile:

That's right that

e

in your formula is the charge of an electron, but it's the charge of a proton as well.

I understand, thankfully :smile:

Can i ask you for help on one more part, it's just i'm not too sure with special relativity.

Question -
Protons are accelerated from rest to a speed, v through a pd of 2900MV.
Show that the mass of a proton at this speed is 4.1×rest mass of a proton.
Answer -
I did eV = Kinetic energy
I found the speed to be 7.45×10^8
But when i substitute it into the special rel equation for mass, i get a -ve value in the square root.

Do you know what to do?
Much apprecited!

PS - Congratulations on your university offers! :smile:

Reply 3

Prime Suspect

Original post by Churchgirl

I understand, thankfully :smile:

m = \gamma m_0

Find v, use that to find gamma and you're done.

Reply 4

Churchgirl

Original post by Prime Suspect

m = \gamma m_0

Find v, use that to find gamma and you're done.

I'm not too sure if i've even come across that formula before O.o

(edited 13 years ago)

Reply 5

Prime Suspect

Original post by Churchgirl

I'm not too sure if i've even come across that formula before O.o

Derivation and some other stuff here

Something's wrong in your case though as the velocity you've got is greater than c?

(edited 13 years ago)

Reply 6

Churchgirl

Original post by Prime Suspect

Derivation and some other stuff here

Something's wrong in your case though as the velocity you've got is greater than c?

All the forrmulae i can use are in:
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-PHA8W-W-QP-JUN07.PDF
On page 4

Reply 7

Prime Suspect

Original post by Churchgirl

All the forrmulae i can use are in:
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-PHA8W-W-QP-JUN07.PDF
On page 4

That formula is on page 4, just written as:

E = mc^2 = \dfrac{m_0c^2}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}

I think though that the problem is that you've got a velocity greater than c - have you checked your numbers?

Reply 8

Churchgirl

Well to find v -
eV = Kinetic energy
Rearranged :
v = (2eV / m)^0.5
2(1.6× 10^-19)(2900^6)/(1.67× 10^-27) then square rooted the whole thing
I get : 7.45 10^8

I can't see where my mistake is but i'm sure i've made one

Reply 9

Prime Suspect

Original post by Churchgirl

No that is correct; I think I know what the problem is though - you're working in the lab frame (as opposed to the particle's rest frame) when making this calculation, so you can't use the rest mass of the proton in this equation when you find kinetic energy

I think you need to say that

E = m_0c^2 + eV

i.e. the total energy E is equal to the sum of the rest mass energy m_0c^2 and the kinetic energy (which we know is equal to eV).

You then use that the total energy E is also given by

E = \gamma m_0 c^2

from your formula book, so putting together:

\gamma m_0 c^2 = m_0c^2 + eV

i.e.

\gamma = 1 + \dfrac{eV}{m_0c^2}

Work this out and I think you should have your answer

(edited 13 years ago)

Reply 10

Churchgirl

Original post by Prime Suspect

E = m_0c^2 + eV

i.e. the total energy E is equal to the sum of the rest mass energy m_0c^2 and the kinetic energy (which we know is equal to eV).

You then use that the total energy E is also given by