Electric Potential Difference

[Physics345]

Homework Statement

In a TV tube, an electric potential difference accelerates electrons from a rest position towards a screen. Just before striking the screen, the electrons have a wavelength of 10×〖10〗^(-11) m. Find the electric potential difference.

Homework Equations

∆E_K=q∆V
λ=h/mv

The Attempt at a Solution

given: e=1.6×10^(-19) Js λ=1.0×10^(-11) m m=9.11×10^(-31) kg h=6.63×10^(-34) Js
λ=h/mv
v=h/λm
v=(6.63×10^(-34) Js )/((1.0×10^(-11) m)(9.11×10^(-31) kg))
v=7.3×10^7 m⁄s
∆E_K=q∆V
E_K2-E_K1=e∆V Rest position so E_K1=0
∆V=(1/2 mv^2)/e
∆V=(1/2 (9.11×10^(-31) kg) (7.3×10^7 m⁄s)^2)/(1.6×10^(-19) Js)
∆V=1.52×10^4 V Therefore the electric potential difference is 1.52×10^4 V

 

[drvrm]

In a TV tube, an electric potential difference accelerates electrons from a rest position towards a screen. Just before striking the screen, the electrons have a wavelength of 10×〖10〗^(-11) m. Find the electric potential difference.

an error in data handling ..the wavelength given is 10x 1o^-11 m. and you have used 1x 10^-11 so it will come to about some different figure for the PD.
just check...

given: e=1.6×10^(-19) Js λ=1.0×10^(-11) m m=9.11×10^(-31) kg h=6.63×10^(-34) Js
λ=h/mv
v=h/λm
v=(6.63×10^(-34) Js )/((1.0×10^(-11) m)(9.11×10^(-31) kg))
v=7.3×10^7 m⁄s
∆E_K=q∆V
E_K2-E_K1=e∆V Rest position so E_K1=0
∆V=(1/2 mv^2)/e
∆V=(1/2 (9.11×10^(-31) kg) (7.3×10^7 m⁄s)^2)/(1.6×10^(-19) Js)
∆V=1.52×10^4 V

moreover a neat way would have been to take wavelength= h/ p the momentum
and write K E = p^2/2m = e. PD...an easier calculation..

 

[Physics345]

Nice catch so basically p=h/wavelength
p=6.63x10^-34/10x10^-11
p=6.63x10^-24
K E= (6.63x10^-24)^2/2(9.11×10^(-31))
K E=4.39569x10^-47/1.822x10^-30
K E=2.412563117x10^-17
Therefore the electric potential difference is 2.412563117x10^-17 eV?

Using
λ=h/mv
v=h/λm
v=(6.63×10^(-34) Js )/((1.0×10^(-11) m)(9.11×10^(-31) kg))
v=7.28×10^22 m⁄s
∆E_K=q∆V
E_K2-E_K1=e∆V Rest position so E_K1=0
∆V=(1/2 mv^2)/e
∆V=(1/2 (9.11×10^(-31) kg) (7.28×10^22 m⁄s)^2)/(1.6×10^(-19) Js)
∆V=1.51x10^31 kgm2/s2 I get this number :S which is correct?

 

[drvrm]

Nice catch so basically p=h/wavelength
p=6.63x10^-34/10x10^-11
p=6.63x10^-24
K E= (6.63x10^-24)^2/2(9.11×10^(-31))
K E=4.39569x10^-47/1.822x10^-30
K E=2.412563117x10^-17
Therefore the electric potential difference is 2.412563117x10^-17 eV?

when you get K.E. , this K E is generated by e. potential difference, so potential difference= K.E./e (charge of an electron) ,calculate and see.

 

[Physics345]

when you get K.E. , this K E is generated by e. potential difference, so potential difference= K.E./e (charge of an electron) ,calculate and see.

p=6.63x10^-34/10x10^-11
p=6.63x10^-24
K E= (6.63x10^-24)^2/2(9.11×10^(-31))
K E=4.39569x10^-47/1.822x10^-30
K E=2.412563117x10^-17/1.6×10^(-19) Js
potential difference=2.412563117x10^-17/1.6×10^(-19) Js
potential difference=1.51x10^2 V
Therefore the electric potential difference is 1.51x10^2 volts?
IT certainly seems correct.

 

[Physics345]

Thanks a lot Drvrm, your help is very much appreciated sir.

 

[Physics345]

p=6.63x10^-34/10x10^-11
p=6.63x10^-24
K E= (6.63x10^-24)^2/2(9.11×10^(-31))
K E=4.39569x10^-47/1.822x10^-30
K E=2.412563117x10^-17
potential difference=2.412563117x10^-17/1.6×10^(-19) Js
potential difference=1.51x10^2 V
Therefore the electric potential difference is 1.51x10^2 volts?
IT certainly seems correct.

I noticed a mistake here the calculations should be the following.
K E=4.39569x10^-47/1.822x10^-29
K E=2.412563117x10^-18
potential difference=2.412563117x10^-18/1.6×10^(-19) Js
potential difference=1.51x10^2 V
Therefore the electric potential difference is 1.51x10 volts?

 

[drvrm]

I noticed a mistake here the calculations should be the following.
K E=4.39569x10^-47/1.822x10^-29
K E=2.412563117x10^-18
potential difference=2.412563117x10^-18/1.6×10^(-19) Js
potential difference=1.51x10^2 V
Therefore the electric potential difference is 1.51x10 volts?

check again...your earlier number seems to be good.

 

[Physics345]

check again...your earlier number seems to be good.

You're correct I made a mistake on paper. Thanks again :)

 

[Physics345]

Well finding the Electric potential difference this way made me me get 0 marks. My original way was correct according to the teacher.

 

[gneill]

The methods are essentially equivalent and should reach the same answer. Penalizing the method used, unless it's been specifically prescribed by the question, seems rather unfair.

The only issue I've seen in this discussion is the specification of the wavelength which changes from $\lambda = 10 \times 10^{-11}~m$ in the problem statement to $1.0 \times 10^-{11}~m$ in the solution attempt.

As a point of interest, as I recall the typical anode potential in a (now old-style!) black&White TV was between 10 to 20 kV, so a 15 kV result seems appropriate. Color TV tubes typically had a larger potential difference, up to around 35 kV. There was a risk of some X-ray exposure close to the surface of the screen before the tube glass was modified to absorb them

 

[Physics345]

The methods are essentially equivalent and should reach the same answer. Penalizing the method used, unless it's been specifically prescribed by the question, seems rather unfair.

The only issue I've seen in this discussion is the specification of the wavelength which changes from $\lambda = 10 \times 10^{-11}~m$ in the problem statement to $1.0 \times 10^-{11}~m$ in the solution attempt.

As a point of interest, as I recall the typical anode potential in a (now old-style!) black&White TV was between 10 to 20 kV, so a 15 kV result seems appropriate. Color TV tubes typically had a larger potential difference, up to around 35 kV. There was a risk of some X-ray exposure close to the surface of the screen before the tube glass was modified to absorb them

I honestly believe it was unfair as well, it's sad though I would have had grade of 100% before heading into the exam if I got this correct. It's okay though this is a minuscule difference, compared to 100%. it is approximately around 99% going into the exam now. I might appeal the mark, considering it isn't technically wrong, basing my appeal off your argument that it does not state which method should be used, so basically it is an issue with the question more than the answer.

 

[Physics345]

Okay I sent out an appeal yesterday my teacher said
"I believe you have the incorrect answer, and that is why you did not earn full marks. I’m having difficulty following your solution. It looks like you found the momentum, then substituted it into a formula, at which point I have difficulty following your logic. You come up with a similar answer to the correct solution, but the kinetic energy in the solution should be 2.4x10-15, and the potential is 1.5x10^4 eV. If you could break down your logic a bit more, I can review your solution further. Try working it out with the v=h/mλ formula followed by E=1/2mv2 and see how the answers differ."
so I guess it is incorrect. I have no way to argue that because when I did the calculations that way I came to the same answer. What should I do now?

 

[drvrm]

so I guess it is incorrect. I have no way to argue that because when I did the calculations that way I came to the same answer. What should I do now?

give a copy of the original question then it can be checked whether wavelength data is 'correct' as your teacher asked. then i think it can get cleared...why not try wavelength value as 10^-11 m...then you may get to your teacher's answer.

 

[Physics345]

λ=h/mv
v=h/λm
v=(6.63×10^(-34) Js )/((10×10^-11m)(9.11×10^-31 kg))
v=7277716.8 m/s
E=1/2(9.11X10^-31 kg)(7277716.8 m/s)^2
E=2.4 x 10^-19 kgm/s
Okay now I am really confused I just did the calculations again... I keep getting E= 2.4 x 10^-19 kgm/s what exactly am I doing wrong here?

 

[drvrm]

E=1/2(9.11X10^-31 kg)(7277716.8 m/s)

have you squared the velocity?

 

[Physics345]

have you squared the velocity?

Yeah I did on paper. opps i made a slight error, but regardless it still doesn't add up
im going to break down my work step by step
λ=h/mv
v=h/λm
v=(6.63×10^(-34) J/s )/((10×10^-11m)(9.11×10^-31 kg))
v=(6.63×10^(-34) J/s )/(9.11x10^-41 kgm/s)
v=7277716.8 kgm/j
E=1/2(9.11X10^-31 kg)(7277716.8 kgm/j)^2
E=1/2(9.11x10^-31 kg)(5.3x10^13 kgm/j)
E= 1/2(4.8283x10^-17)
E=2.4 x 10^-17
on a side note is kgm/j written like that I forget.

 

[drvrm]

Yeah I did on paper.

its troubling you..the arithmatic..so check it afresh

If you take lambda = h/mv then v =h/(m. Lambda)...and V^2 = (h^2)/(m.lambda)^2

so K.E=( 1/2) m.{ (h^2) /(m*2.lambda^2)}

so the potential diff. V= {(h^2) /(2m e)} x (1/ lambda^2)

the factor is of the order of 10^-18 / Lambda ^2 so if wavelength is of the order of 10^-11 you can get V of 10^4

 

[Physics345]

V^2 = (h^2)/(m.lambda)^2 what is this part for? and m.lambda means mass x lambda correct?

 

[drvrm]

V^2 = (h^2)/(m.lambda)^2 what is this part for? and m.lambda means mass x lambda correct?

i.e square of the velocity v^2 so that KE can be calculated, m is mass

 

[Physics345]

K.E=( 1/2) m.{ (h^2) /(m*2.lambda^2)}
= (1/2)9.11×10^-31[(6.63×10^-34)^2/2(9.11×10^-31)(10×10^-11)^2]
?
v=sqrt(6.63x10^-34)^2/(9.11x10-31.10x10-11)^2
?
The main issue here is I haven't touched physics for a while I need to review.

 

[drvrm]

K.E=( 1/2) m.{ (h^2) /(m*2.lambda^2)}
= (1/2)9.11×10^-31[(6.63×10^-34)^2/2(9.11×10^-31)(10×10^-11)^2]
?

so one m get canceled K.E.= (1/2).(h^2/ (m x lambda^2) and you divide by e ,then you get a factor/lambda^2

 

[Physics345]

so one m get canceled K.E.= (1/2).(h^2/ (m x lambda^2) and you divide by e ,then you get a factor/lambda^2

so after those calculations I divide by e?
so 2.4x10^-17/e?
KE=2.4x10^-17/1.6x10^-17
KE=15
if this is the case where is my teacher getting his numbers from? I am basically getting the same answer I did with your momentum method.

 

[drvrm]

so after those calculations I divide by e?
so 2.4x10^-17/e?

no i am saying that the P.D. = A factor x(1/ lambda^2) that
factor is (h^2)/ (2.m.e) so you can test what wavelength should need which Pot. Diff. and can check the teacher's view.

 

[Physics345]

no i am saying that the P.D. = A factor x(1/ lambda^2) that
factor is (h^2)/ (2.m.e) so you can test what wavelength should need which Pot. Diff. and can check the teacher's view.

okay so factor=1.51x10^-20
now
P.D= 1.5x10^-20 x (1/(10x10^-11)^2)
=1.5x10^-20 x (1/1x10-20)
=1.5x10^-20(1x10^20)
=1.5x10
=15
:S there is something I'm not understanding here clearly.

 

[drvrm]

okay so factor=1.51x10^-20
now
P.D= 1.5x10^-20 x (1/(10x10^-11)^2)
=1.5x10^-20 x (1/1x10-20)
=1.5x10^-20(1x10^20)
=1.5x10
=15
:S there is something I'm not understanding here clearly.

i get the factor as 10^-18 ,check the numbers.

 

[Physics345]

i get the factor as 10^-18 ,check the numbers.

after doing it again I am getting the following:
4.4x10^-67/2[(9.11x10^-31)(1.6x10^-17)]
4.4x10^-67/2.92x10-47
=1.3x10^-19

 

[drvrm]

1.6x10^-17)]

what is this number doing there

 

[Physics345]

Oops, that should be 1.6x10^-19.. that's the mistake I guess.

 

[Physics345]

4.4x10^-67/2[(9.11x10^-31)(1.6x10^-17)]
4.4x10^-67/2.92x10-47
=1.5x10^-18
P.D= 1.5x10^-18 x (1/(10x10^-11)^2)
=1.5x10^-18 x (1/1x10-20)
=1.5x10^-18(1x10^20)
=1.5x10^2
=150

 

[drvrm]

4.4x10^-67/2[(9.11x10^-31)(1.6x10^-17)]
4.4x10^-67/2.92x10-47
=1.5x10^-18
P.D= 1.5x10^-18 x (1/(10x10^-11)^2)
=1.5x10^-18 x (1/1x10-20)
=1.5x10^-18(1x10^20)
=1.5x10^2
=150

that's why i an of the opinion that wavelength must be corrected

 

[Physics345]

thats why i an of the opinion that wavelength must be corrected

What do you mean? by "I an of the opinion must be correct"

 

[drvrm]

What do you mean? by "I an of the opinion must be correct"

I am referring to your teacher's reply which you wrote earlier that the P.D. should be !0^4 order ...that's why i am saying that unless wavelength gets changed to lower value the P.D. can not increase.. if you try only !0^-11 as wavelength you can get that order of P.D.

 

[Physics345]

so basically, the wavelength won't allow for any higher P.D. than we are getting here. Basically if I list the work we just did it should explain to him why my answer is correct?

 

[Physics345]

This is really silly though, I shouldn't have to explain this to him. He should of done the math him self and came to the conclusion my answer was correct. Praise be to drvrm! I appreciate you helping me teach the teacher. This feels great =) I asked for extra marks for the explanation.