Finding Electric Potential Difference in a TV Tube

[julianwitkowski]

Homework Statement

In a TV tube, an electric potential difference accelerates electrons from a rest positive towards a screen. Just before striking the screen, the electrons have a wavelength of 1.0⋅10^-10. Find the electric potential difference.

Homework Equations

[/B]
Should I use these?

$$\Delta V = \frac{Work}{Charge}$$

$$λ= \frac{h}{\sqrt{2⋅m⋅E}}$$

The Attempt at a Solution

substituting/simplifying I came to: $$\Delta V = \frac{h^2}{2⋅e⋅m⋅λ^2}$$

Would this best the correct equation to find the answer?

***Corrected the 'E' implying energy to 'e' for elementary charge

Thanks for your time and help,
Julian

 

[gneill]

In a TV tube, an electric potential difference accelerates electrons from a rest positive towards a screen. Just before striking the screen, the electrons have a wavelength of 1.0⋅10^-10. Find the electric potential difference.

1.0⋅10^-10 what? Miles? Astronomical units? Watermelons?

Be sure to include units for every numerical data value!

The equation that you arrived at looks to be okay. Not sure how you got there from your Relevant Equations though

 

[julianwitkowski]

Sorry about my units...

Since $$λ= \frac{h}{\sqrt{2⋅m⋅E}} → E= \frac{h^2}{\sqrt{2⋅m⋅λ}}$$

I reckon: $$ΔV = \frac{E}{e} = \frac{(\frac{h^2}{\sqrt{2⋅m⋅λ}})}{e} = ΔV = \frac{h^2}{\sqrt{2⋅m⋅e⋅λ}}$$

 

[julianwitkowski]

1.0⋅10^-10

The equation that you arrived at looks to be okay. Not sure how you got there from your Relevant Equations though

 

[julianwitkowski]

I was looking at this and my second algebra is terrible... I meant to write this but I was awake for way too long.Since $$λ= \frac{h}{\sqrt{2⋅m⋅E}} → E= \frac{h^2}{2⋅m⋅λ^2}$$

$$ΔV = \frac{E}{e} = \frac{(\frac{h^2}{2⋅m⋅λ^2})}{e} = \frac{h^2}{2⋅m⋅e⋅λ^2}$$

 

[gneill]

Good. You've shown your work and it looks fine to me

Now, what are the units of the wavelength you've been given? It matters because it will determine the magnitude of the velocity of the electron; you need to be sure that the speed is low enough that relativistic equations aren't required (as it turns out they aren't here, but it is something you should check for every time).

 

[julianwitkowski]

Good. You've shown your work and it looks fine to me

Now, what are the units of the wavelength you've been given? It matters because it will determine the magnitude of the velocity of the electron; you need to be sure that the speed is low enough that relativistic equations aren't required (as it turns out they aren't here, but it is something you should check for every time).

It's meters, and thanks for your help I mostly needed to make sure it was right before I put in the numbers; that's the easy part :p.

 

[julianwitkowski]

Good. You've shown your work and it looks fine to me

Now, what are the units of the wavelength you've been given? It matters because it will determine the magnitude of the velocity of the electron; you need to be sure that the speed is low enough that relativistic equations aren't required (as it turns out they aren't here, but it is something you should check for every time).

Question about significant digits... Is this one significant digit because the question is 1.0⋅10^-10m?

 

[BvU]

I wouldn't worry and quote 150 V: If someone gives 1.0 then in principle you are to assume it's in the range 0.95 - 1.05, so 1.5 10² V would be a good answer.

Sure, the numbers are the easy part. If you found 150 V like I did (did you ?) I suppose that's what the exercise wants.

(But that is a funny kind of TV: I seem to remember a color TV has some 25 kV as accelerating voltage. Perhaps they wanted to make sure you wouldn't worry about relativistic corrections... )

--

 

[julianwitkowski]

I wouldn't worry and quote 150 V: If someone gives 1.0 then in principle you are to assume it's in the range 0.95 - 1.05, so 1.5 10² V would be a good answer.

Sure, the numbers are the easy part. If you found 150 V like I did (did you ?) I suppose that's what the exercise wants.

(But that is a funny kind of TV: I seem to remember a color TV has some 25 kV as accelerating voltage. Perhaps they wanted to make sure you wouldn't worry about relativistic corrections... )

--

...and wait, isn't electric potential diff in eV.. I jump at the conclusion due to the ΔV symbol and as eV's seem to be standard in this study? ... Thanks for pointing that out.

 

[gneill]

...and wait, isn't electric potential diff in eV.. I jump at the conclusion due to the ΔV symbol and as eV's seem to be standard in this study? ... Thanks for pointing that out.

Electric potential difference has units of Volts, which is equivalent to Joules per Coulomb. eV is a unit of energy alone, i.e. some number of Joules.

 

[julianwitkowski]

...and wait, isn't electric potential diff in eV.. I jump at the conclusion due to the ΔV symbol and as eV's seem to be standard in this study? ... Thanks for pointing that out.

I need to learn to read still...

 

[julianwitkowski]

BvU, you're honestly a tank. Thanks dude.

 

[julianwitkowski]

I wouldn't worry and quote 150 V: If someone gives 1.0 then in principle you are to assume it's in the range 0.95 - 1.05, so 1.5 10² V would be a good answer.
--

But why the heck does the equation render a value within the realm of the centi-volt? I pull 15078.5 from the equation but I'm not sure the units,

and if you say its 150 volts, I'm in a new place of confusion.

 

[julianwitkowski]

Thank you for the critical thinking exercise.

 

[gneill]

But why the heck does the equation render a value within the realm of the centi-volt? I pull 15078.5 from the equation but I'm not sure the units,

Start by writing out the values of all the variables and constants that you are using in the equation. Make sure that they all have compatible units (i.e. don't mix meters and centimeters and so on). You will be able to find the final units if you combine/cancel them as your calculation proceeds.

 

[julianwitkowski]

Start by writing out the values of all the variables and constants that you are using in the equation. Make sure that they all have compatible units (i.e. don't mix meters and centimeters and so on). You will be able to find the final units if you combine/cancel them as your calculation proceeds.

I might be doing something wrong here, I tried this a couple of ways but I don't exactly line up with $$\frac{kg⋅m^2}{A⋅s^3}$$

I end up with this

$$\frac{m}{A⋅s^2}$$

 

[julianwitkowski]

I'm not sure if you can see where I'm going wrong in my units, I've tried using Plancks constant with J/s too but its also not Volts, although I know it is, but where am I going wrong.

 

[gneill]

I'm not sure if you can see where I'm going wrong in my units, I've tried using Plancks constant with J/s too but its also not Volts, although I know it is, but where am I going wrong.

Please post the values with units. I can tell you right now that the units of h are J s or kg m²/s, not J/s.

 

[julianwitkowski]

I forgot to square the Planck constant and wavelength when working with the units alone... Were all good, thanks for the tip, I thought it might be easier to think about working with the units without numbers but I forgot about the powers of the value to begin.. I certainly see the importance of the actual units now, so I really appreciate everyone here drilling that stuff for me. I love this site.