is λ=h/mv the same equation as λ=hc/E ?

that was the answer i was looking forward to but the problem is that my book says that a beam of moving electrons has associated de broglie wavelength (where at high speed you have to take into account relativistic effects) is approx λ≃hc/E can you please tell me what does that mean?

The equation for p, in relativity is:

$E^2 = p^2c^2 + m^2c^4$

Just saying, $m$ is the rest mass and usually denoted by $m_0$ because in relativity, relativistic mass $m$ is related to the rest mass by $m=\gamma m_0$.

My relativity lecturer used to say that the usage of "relativistic mass" was a waste of time that needlessly complicated the issue. Just saying.



(After all, for a particle and setting c = 1,

$E = \gamma m$

And so relativistic mass and energy are essentially the same thing. Why complicate the situation with different notions of mass? )

that was the answer i was looking forward to but the problem is that my book says that a beam of moving electrons has associated de broglie wavelength (where at high speed you have to take into account relativistic effects) is approx λ≃hc/E can you please tell me what does that mean?

It's not a mistake. If a beam of electrons is moving extremely fast - at almost the speed of light, the vast majority of its energy will be due to its momentum, not its rest mass. So we can say:

$E^2 = p^2c^2 + m^2c^4 \simeq p^2c^2$

and so we get

$E \simeq pc$

As a good approximation.

Look again at bbcs2k9's post above. E=hf (or E=hc/λ, they are the same equation) is for photons. If you wnat to find the deBroglie wavelength of a particle, you need to use λ=h/mv.

my book says that a beam of moving electrons has associated de broglie wavelength (where at high speed you have to take into account relativistic effects) is approx λ≃hc/E
so is it wrong?

so to get this straight there are two situations:
1) if the particle is moving with slow speed we cant use λ=hc/E since this is only applicable to photons and in this case we only use λ=h/mv
2) if the particle is moving with very high speed we can use λ=hc/E and not λ=h/mv
(+ can you explain to me why again in a simpler way )

That's correct. Not sure how else to explain it. But basically, λ=hc/E is valid for a particle moving at the speed of light. But if a particle is moving very close to the speed of light, to a very good approximation p = E/c and you can still write λ=hc/E. However, if the particle is moving nowhere near the speed of light you need to use the normal expression for momentum p = mv, instead of p = E/c and so you get λ=h/mv .

That's correct. Not sure how else to explain it. But basically, λ=hc/E is valid for a particle moving at the speed of light. But if a particle is moving very close to the speed of light, to a very good approximation p = E/c and you can still write λ=hc/E. However, if the particle is moving nowhere near the speed of light you need to use the normal expression for momentum p = mv, instead of p = E/c and so you get λ=h/mv .

but we cant use λ=h/mv for particles moving close to speed of light right one can only use λ=hc/E right?
and thanks

but we cant use λ=h/mv for particles moving close to speed of light right one can only use λ=hc/E right?
and thanks

We use $\lambda=\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}$ when a particle with mass moving in a speed comparable to the speed of light.

We use $\lambda=\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}$ when a particle with mass moving in a speed comparable to the speed of light.

As Agostino says:




So when v = 0,


$\lambda=\frac{h}{mv}\sqrt{1-\frac{0^2}{c^2}} = \lambda/mv$

The expression for momentum that you're familiar with.

At A Level, which equation to use is very simple.

λ=hc/E (or E=hc/λ, or E=hf, they are all differenct versions of the same equation) is for photons.

λ=h/mv is for matter (i.e. anything with mass).

Strictly speaking, you need to adapt λ=h/mv when v is close to c, because of relativity (and some here have described how). This will not be expected of you at A Level.

As Agostino says:

So when v = 0,


$\lambda=\frac{h}{mv}\sqrt{1-\frac{0^2}{c^2}} = \frac{h}{mv}$

The expression for momentum that you're familiar with.

ok so once and for all λ=h/mv-------> particles but needs adjustments when close to speed of light
& λ=hc/E --------->photons in A level right?

Nope. Please note that if v=0, you get $\lambda=\frac{h}{m(0)}$. What you have is when $v\ll c$, you have $\sqrt{1-\frac{v^2}{c^2}}\approx 1$, thus you have $p=mv$, which is the equation you are familiar with, which is $\lambda=\frac{h}{mv}$.