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

if not when is each used?

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ibs2012

Someone correct me if I'm wrong, but from what I remember:

λ=h/p
from mechanics: p=mv
so: λ=h/mv
Photons move at the speed of light, so v=c
λ=h/mc

At the same time, from Einstein: E=mc^2
m=E/c^2
thus: λ=h/[(E/c^2)c]
λ=h/(E/c)
λ=hc/E

So basically both equations are related, but not quite the same. You need to apply them depending on what variables are known and which ones you are trying to find.

I hope that helps.

Reply 2

agostino981

Original post by ibs2012

Photon is massless, so you can't really apply

p=mv

.

But, what you can do is apply

E=h\nu

and

c=\lambda \nu

, where

\nu

is the frequency and

\lambda

is the wavelength.

EDIT:

Original post by >>MMM<<

i am now confused so they are the same or not if not why and when do we use each?!

Photon is massless, so

m=0

which gives you nothing if you use

p=mv

What I did is to apply the equation for energy level

E=h\nu

and the equation relating the speed of wave, wavelength and frequency

v=f\lambda

, because it is a photon, you have

v=c

, hence

c=\nu \lambda

If you have something with mass, you can, of course, apply

p=mv

p=\gamma mv

for relativistic condition. However, you are considering photon, for momentum, you have

p=\frac{h}{\lambda}

(edited 11 years ago)

Reply 3

>>MMM<<

Original post by agostino981

Photon is massless, so you can't really apply

p=mv

.

But, what you can do is apply

E=h\nu

and

c=\lambda \nu

, where

\nu

is the frequency and

\lambda

is the wavelength.

i am now confused so they are the same or not if not why and when do we use each?!

Reply 4

ibs2012

Original post by agostino981

Photon is massless, so you can't really apply

p=mv

.

But, what you can do is apply

E=h\nu

and

c=\lambda \nu

, where

\nu

is the frequency and

\lambda

is the wavelength.

Ah yes, of course, thanks for the correction!

Reply 5

ibs2012

Original post by >>MMM<<

i am now confused so they are the same or not if not why and when do we use each?!

I'm assuming that your variables are m=mass,v=velocity?
In that case the only use I can find for λ=h/mv is when you are trying to find the de Broglie wavelength of larger objects, ie a car or person moving.

Reply 6

>>MMM<<

Original post by ibs2012

how is that possible? i have read in my book that λ=h/mv can be used in any situation? :s-smilie:

Reply 7

agostino981

Original post by >>MMM<<

how is that possible? i have read in my book that λ=h/mv can be used in any situation? :s-smilie:

Except massless or static object.

Reply 8

bbcs2k9

Original post by ibs2012

λ=h/mv is known as the Broglie wavelength and you use it when you are asked to find the associated wavelength of a particle such as an electron with a certain speed.

λ=hc/E is the wavelength of a photon with a certain energy. You usually see the equation in this form: E=hc/λ. and they will usually ask to find the wavelength energy given a certain wavelength.

I dont think they are they same (they might be, but I dont' know). But generally in my course (OCR), λ=h/mv is for particles and E=hc/λ is for photons of light.

(edited 11 years ago)

Reply 9

>>MMM<<

Original post by bbcs2k9

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?

Reply 10

theterminator

Original post by >>MMM<<

If you want to calculate the relativistic de broglie wavelength of a particle of rest mass

m_{0}

then the equation becomes:

Unparseable latex formula:

\displaystyle {\lambda}_{de-Broglie} = \frac{h}{vcm_{0}} \displaystyle \sqrt_{c^2 - v^2}

(edited 11 years ago)

Reply 11

>>MMM<<

Original post by theterminator

If you want to calculate the relativistic de broglie wavelength of a particle of rest mass

m_{0}

then the equation becomes:

Unparseable latex formula:

\displaystyle {\lambda}_{de-Broglie} = \frac{h}{vcm_{0}} \displaystyle \sqrt_{c^2 - v^2}

thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!

Reply 12

theterminator

Original post by >>MMM<<

thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!

I think you may be getting confused with the derivation of the de Broglie equation from above. The above derivation is the one taught but isn't actually the correct way of deriving the de Broglie wavelength. The correct derivation requires more fundamental special relativity but it makes the clear distinction between photons and electrons. Something you should know about the de Broglie wavelength is that it is a relativistic effect, and E=mc^2 is an equation for something that has zero velocity in the frame of reference being used.

Reply 13

Pangol

Original post by >>MMM<<

thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!

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.

Reply 14

agostino981

Original post by >>MMM<<

thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!

Nope,

\lambda = \frac{hc}{E}

is for photons only.

If you want to find the de-Broglie wavelength of electrons, you need

\lambda=\frac{h}{p}

.

If you seek a non-relativistic result, you apply

p=mv

, if you seek a relativistic result, you apply

p=\gamma m v

, where

\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

A pretty obvious reason why you can't use

\lambda=\frac{hc}{E}

is because electrons cannot reach light speed.

Reply 15

>>MMM<<

Original post by agostino981

Nope,

\lambda = \frac{hc}{E}

is for photons only.

If you want to find the de-Broglie wavelength of electrons, you need

\lambda=\frac{h}{p}

.

If you seek a non-relativistic result, you apply

p=mv

, if you seek a relativistic result, you apply

p=\gamma m v

, where

\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

A pretty obvious reason why you can't use

\lambda=\frac{hc}{E}

is because electrons cannot reach light speed.

so your basically saying λ=h/mv is only for particles and λ=hc/E only for photons and the book has some kind of mistake in it?

Reply 16

>>MMM<<

Original post by Pangol

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.