# How Is Electric Potential Difference Calculated in a TV Tube?

• Physics345
In summary: Thanks for the help :)In summary, the conversation discussed the calculation of electric potential difference in a TV tube, where an electric potential accelerates electrons towards a screen. The given wavelength of the electrons was 10x10^-11 meters, and using the formula λ=h/mv, a velocity of 7.3x10^7 m/s was calculated. The formula ∆E_K=q∆V was then used to find the potential difference, resulting in a value of 1.51x10^2 volts. The potential difference in a TV tube is typically between 10 to 35 kV, making this result reasonable. However, there was a discrepancy in the given wavelength, which could affect the
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.

∆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

Physics345 said:
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...
Physics345 said:
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?

Last edited:
drvrm
Physics345 said:
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
drvrm said:
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.

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

drvrm
Physics345 said:
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
Physics345 said:
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
drvrm said:
check again...your earlier number seems to be good.
You're correct I made a mistake on paper. Thanks again :)

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

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
gneill said:
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.

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?

Physics345 said:
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?

Last edited:
Physics345 said:
E=1/2(9.11X10^-31 kg)(7277716.8 m/s)

have you squared the velocity?

Physics345
drvrm said:
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.

Last edited:
Physics345 said:
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?

Physics345 said:
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.

Physics345 said:
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
drvrm said:
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.

Physics345 said:
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
drvrm said:
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.

Physics345 said:
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.

drvrm said:
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

Physics345 said:
1.6x10^-17)]

what is this number doing there

Physics345
Oops, that should be 1.6x10^-19.. that's the mistake I guess.

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

Physics345 said:
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

drvrm said:
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"

Physics345 said:
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?

drvrm
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.

drvrm

## 1. What is electric potential difference?

Electric potential difference, also known as voltage, is the difference in electric potential energy per unit charge between two points in an electric field. It is measured in volts (V).

## 2. How is electric potential difference related to electric current?

Electric potential difference is directly related to electric current. The greater the potential difference, the higher the electric current will be. This is because the potential difference causes the flow of charged particles, known as electrons, from a region of higher potential to a region of lower potential.

## 3. How is electric potential difference different from electric potential?

Electric potential difference is a measure of the difference in potential energy between two points, while electric potential is a measure of the potential energy at a single point in an electric field. In other words, electric potential difference is a comparison between two points, while electric potential is the potential energy at a specific point.

## 4. What factors affect electric potential difference?

Electric potential difference is affected by the distance between two points, the amount of charge present, and the strength of the electric field. The greater the distance, the smaller the potential difference. The greater the charge, the greater the potential difference. And the stronger the electric field, the greater the potential difference.

## 5. How is electric potential difference measured?

Electric potential difference is measured using a voltmeter, which is connected in parallel to the circuit. The voltmeter measures the potential difference between two points in a circuit and displays the value in volts. It is important to note that the voltmeter must have a high resistance to avoid altering the potential difference being measured.

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