Wavelength = deBroglie Wavelength?

  • #1
Krushnaraj Pandya
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Homework Statement


Find the retarding potential required to stop electron of de-broglie wavelength 0.5 nm

Homework Equations


1)de-broglie wavelenght=h/momentum
2)eV=KE of electron
3)KE=p^2/2m

The Attempt at a Solution


Using the above three relations is giving me an incorrect answer for V, However in the hint it is mentioned that eV=hc/lambda.
I have four problems- 1)Why are the above relations giving an incorrect answer?
2) hf is the energy of a photon, can we use the same expression for an electron?
3)if yes, How did we obtain this expression?
4)the photon is massless, so is its de-broglie wavelength infinite?
I'd be grateful for your help
 
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Answers and Replies

  • #2
Krushnaraj Pandya
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Ok, so I just solved this using energy conservation. Initially KE=k and and U=0, finally U=eV and KE=0. Therefore k=eV, replacing it in the expression for de-broglie wavelength, I get V=6 volts which is correct
 
  • #3
Ray Vickson
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Homework Statement


Find the retarding potential required to stop electron of de-broglie wavelength 0.5 nm

Homework Equations


1)de-broglie wavelenght=h/momentum
2)eV=KE of electron
3)KE=p^2/2m

The Attempt at a Solution


Using the above three relations is giving me an incorrect answer for V, However in the hint it is mentioned that eV=hc/lambda.
I have four problems- 1)Why are the above relations giving an incorrect answer?
2) hf is the energy of a photon, can we use the same expression for an electron?
3) How did we obtain this expression?
4)the photon is massless, so is its de-broglie wavelength infinite?
(1) We cannnot say, since you do not show your work.
(2) Yes.
(3) Which expression do you mean? You wrote three of them.
(4) Nonsense: the de-Broglie wavelength of a photon is just the wavelength ##\lambda## of the light. Therefore, for a photon, ##p = h \lambda.##

You cannot say that ##\text{KE} = p^2/(2m)## unless you know that the electron travels at speeds considerably less than the speed of light. So, first you need to work out the electron's speed, which you can do by first finding the momentum ##p## then solving the equation
$$\frac{v}{\sqrt{1 - (v/c)^2} }= p/m.$$
So, if ##v## is no more than about 15-20% of ##c##, using the classical expression for ##\text{KE}## will be OK: the errors will be small and not very important. However, if ##v## is a significant fraction of ##c## the errors due to use of the classical vs. relativistic formula will start to matter. Actually, a good exercise is to work out the value or ##\text{KE}## both ways, using the simpler classical formulas ##p = m v## and ##\text{KE} = (1/2) m v^2## and then again using the relativistic formulas ##p = m \gamma v## and ##\text{KE} = m c^2 (\gamma - 1),## where ##\gamma = 1/\sqrt{1-(v/c)^2}.##
 
  • #4
Krushnaraj Pandya
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(1) We cannnot say, since you do not show your work.
Here's my work- 5 x 10^(-10)=de broglie wavelength. Using DBW=h/p, we get p=h/dbw = 6.6 x 10^(-34)/5 x 10^(-10)= 1.32 x 10^(-24). Using p^2/2m=eV, we get V equal to...the correct answer. 6 V. So this was the first, lengthiest method.

(2) Yes.
(3) Which expression do you mean? You wrote three of them.
I meant, if the DBW of an electron is R, how can we write its energy as hc/R. (How did we obtain this expression?)
Also, does this mean I can write hc/R=1/2mv^2 ?
Therefore, for a photon, p=hλ.p=hλ.p = h \lambda.
shouldn't this be h/lambda instead of h into lambda?

The second easier method was energy conservation in post #2 and if I understand correctly, I can solve easily by using hc/lambda =eV also.

Sorry for the late reply, I wanted to understand things properly and then respond accordingly. Thank you for taking the time to help me :D
 
  • #5
Krushnaraj Pandya
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(2) Yes.
This implies an electron and a photon propagating in the form of a wave with same wavelength have equal energy but I found they have equal momentum instead, what's wrong here?
 
  • #6
Krushnaraj Pandya
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There definitely is something wrong, I now understand that since p=h/lambda they have equal momentum for equal wavelengths. However we cannot say they have equal energy (i.e, energy for electron=hf is not valid) I do not know why it is not valid though
 
  • #7
Orodruin
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The relationship between the frequency and the wavelength of an electron is not ##\lambda f = c##.
 
  • #8
Krushnaraj Pandya
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The relationship between the frequency and the wavelength of an electron is not ##\lambda f = c##.
Surprising. I thought that relationship was universal.
 
  • #9
Krushnaraj Pandya
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To what extent is the relation true? and why not for an electron?
 
  • #10
Orodruin
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It only holds for massless particles. The universal statement is ##\lambda f = v_p##, where ##v_p## is the phase velocity. The phase velocity for massive particles is not the speed of light.
 
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  • #11
Krushnaraj Pandya
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Oh! alright, then the relation hf=energy of electron is true but hc/lambda is not. Interesting.
 
  • #12
Krushnaraj Pandya
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Thank you :D
 

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