De broglie wavelength of fast moving particle-help please

[The Head]

De broglie wavelength of fast moving particle--help please!

Homework Statement

A particle moving with kinetic energy equal to its rest energy has a de Broglie wavelength of 1.7898 *10^-16 m. If the kinetic energy doubles, what is the new de Broglie wavelength?

Homework Equations

λ=h/√(2mK)
m=m₀γ
K=E_t-E₀

The Attempt at a Solution

Initially:
K=m₀c²=E_T-m₀c²
But E_T=m₀c²γ
thus: 2=γ, which implies v=1/2(√3)
λ=h/√(2mK)=h/√2m₀γ*m₀c²=
h/(2m₀c)

Then when the K becomes twice the rest mass:
K=E_t-E₀
3m₀c²=m₀c²γ, which implies γ=3
so λ=h/√(2m₀γK)=h/√2m₀*3*(2m₀c²
=h/(2m₀c*√3), which is just the original λ divided by the root of 3.
1.7898/√3= 1.03, so λ=1.033*10^-16, but the answer is λ=1.096*10^-16

Please help-- I have been wracking my brains about this for two days!

 

[collinsmark]

Homework Statement

A particle moving with kinetic energy equal to its rest energy has a de Broglie wavelength of 1.7898 *10^-16 m. If the kinetic energy doubles, what is the new de Broglie wavelength?

Homework Equations

λ=h/√(2mK)
m=m₀γ
K=E_t-E₀

The Attempt at a Solution

Initially:
K=m₀c²=E_T-m₀c²
But E_T=m₀c²γ
thus: 2=γ,

I think I understand your approach. If so, it is a valid way to solve this problem. It's clever to catch on that [itex] \gamma [/itex] = 2 that way (which is correct). (There's more than one way to solve this problem, but this way is as good as any.)

which implies v=1/2(√3)

You're missing a 'c' somewhere. Also, your notation is a little ambiguous.

Do you mean [itex] v = \frac{\sqrt{3}}{2}c [/itex] ?

λ=h/√(2mK)=h/√2m₀γ*m₀c²=
h/(2m₀c)

You totally lost me there. What happened to the [itex] \sqrt{3} [/itex] ?

[tex] \lambda = \frac{h}{p} [/tex]
which is also (since [itex] p = \gamma m_0 v [/itex] ),
[tex] \lambda = \frac{h}{\gamma m_0 v} [/tex]
Since you can figure out [itex] \gamma [/itex] and [itex] v [/itex], then you can solve for [itex] m_0 [/itex] (or at least develop an expression for it as a function of the original [itex] \lambda [/itex]).

Then when the K becomes twice the rest mass:
K=E_t-E₀
3m₀c²=m₀c²γ, which implies γ=3

Yes, [itex] \gamma [/itex] = 3 this time.

so λ=h/√(2m₀γK)=h/√2m₀*3*(2m₀c²
=h/(2m₀c*√3), which is just the original λ divided by the root of 3.
1.7898/√3= 1.03, so λ=1.033*10^-16,

Once again, I'm not following you. You've found the new [itex] \gamma [/itex], which is 3, but how did you calculate the new velocity, [itex] v [/itex] ? It's not clear to me how you did that.

but the answer is λ=1.096*10^-16

Yes, that is the correct answer. (So it's not a mistake in the book or anything like that for this one.)

 

[The Head]

Thank you so much! Yes, somehow I dropped the √3/2 term and never realized. I ended up using the relation λ=h/p. Don't know why I didn't use that in the first place, but it made it much easier.

I have a question though. Using the relation above, I saw that h/p=h/m₀γv, so in the first case with γ=2 you get λ=h/p=h/m₀2*√3/2*c.

If you use,the equation I was originally working with you have λ=h/√(2mK). Now m=m₀γ, and K=m₀c^2(y-1),

so you end up with λ=h/√(2m₀γ*m₀c^2(y-1))= h/(m₀c*√(2γ(γ-1)), where the argument of the square root equals 4 for gamma=2,

and so the expressions are almost the same, except one has a factor of √(3)/2 in the denominator and the other has a √4=2 in the denominator. So I've done something wrong when I try and solve it using the relation with kinetic energy. Any thoughts?

 

[collinsmark]

Thank you so much! Yes, somehow I dropped the √3/2 term and never realized. I ended up using the relation λ=h/p. Don't know why I didn't use that in the first place, but it made it much easier.

I have a question though. Using the relation above, I saw that h/p=h/m₀γv, so in the first case with γ=2 you get λ=h/p=h/m₀2*√3/2*c.

If you use,the equation I was originally working with you have λ=h/√(2mK). Now m=m₀γ, and K=m₀c^2(y-1),

so you end up with λ=h/√(2m₀γ*m₀c^2(y-1))= h/(m₀c*√(2γ(γ-1)), where the argument of the square root equals 4 for gamma=2,

and so the expressions are almost the same, except one has a factor of √(3)/2 in the denominator and the other has a √4=2 in the denominator. So I've done something wrong when I try and solve it using the relation with kinetic energy. Any thoughts?

Okay, I think I see what's happening now. I was previously having so much trouble following your work because I couldn't figure out where the λ=h/√(2m₀K) relationship came from. (I've never used that equation before, myself.)

I finally figured out that [itex] p = \sqrt{2m_0 K} [/itex] is only an approximation, and it's only relevant a non-relativistic speeds, [itex] v \ll c [/itex] (when the kinetic energy is negligible compared to the rest-mass energy).

See here for details: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/debrog2.html#c2.

The gist of it is that at relativistic energies, [itex] p = \sqrt{2m_0 K} [/itex] is not valid (and that's why things weren't working out for you). Instead, you'd need to use [itex] p = \frac{\sqrt{K^2 + 2m_0 Kc^2}}{c} [/itex]

The approximation comes in when K is very small compared to m₀c². Then the K² term can be ignored, and it the approximation for p comes out to be what you were using. But you can't do that for this problem since the kinetic energy is really large. It's equal to the rest mass energy in the first case, and double the rest mass energy in the second. (Definitely relativistic.)

On the other hand, even without using the approximation, it can be shown that [itex] p = \frac{\sqrt{K^2 + 2m_0 Kc^2}}{c} = \gamma m_0 v^2[/itex] which is much easier to work with. (Well, whether it's easier or not for this particular problem is debatable. But as I mentioned before, there are a number of ways to do this problem.)

 

[The Head]

Okay, that makes a lot more sense now! I definitely didn't realize the limits on that equation. I really appreciate you taking the time to help me with this problem-- thank you!