I am new here, have a physics related question, is this the right area?

[kjf]

Hey everyone,

 

First off - I've been on this forum for about 5 minutes now and it's amazing. So much knowledge, I love it.

 

Second - Because I'm new here I don't know if this is the right section for me to post a question?

 

This is the question:

 

"An old-fashioned television works by accelerating electrons through a voltage of 32,000V and onto a screen. Find the velocity and the associated deBrogile wavelength of the electrons as they hit the screen."

 

I gave it a shot, here's what I did.

 

 

K = eV = 32,000V x 1.6x10^-19 C

 

1/2 mv^2 = 5.12 x 10^-15 Js

 

Solving gave me an astounding velocity of 3.35 x 10^9 m/s. Which is impossible, literally.

 

Was just wondering if anyone could give me a nudge in the right direction?

 

Thanks!

:shrug:

 

 

 

 

edit: heh, on a side note - why does zinc dissolve well in copper? (another question that needs answering!)

[CraigD]

First off - I've been on this forum for about 5 minutes now and it's amazing. So much knowledge, I love it.

Welcome to hypography kjf, and thanks for the kind words. :)

 

Feel free to start a thread in the introductions forum, to tell us all a bit about you personally. We like to be a bit personal and social here – though only if you want to.

"An old-fashioned television works by accelerating electrons through a voltage of 32,000V and onto a screen. Find the velocity and the associated deBrogile wavelength of the electrons as they hit the screen."

 

I gave it a shot, here's what I did.

 

 

K = eV = 32,000V x 1.6x10^-19 C

 

1/2 mv^2 = 5.12 x 10^-15 Js

 

Solving gave me an astounding velocity of 3.35 x 10^9 m/s. Which is impossible, literally.

 

Was just wondering if anyone could give me a nudge in the right direction?

When Newton gives you the impossible, use relativity!

 

The energy of an electron at rest is about 510998.91 eV.

 

Adding 32000 eV of energy to it by subjecting it to an electrostatic difference of 32000 V gives it an energy of about 542998.91 eV.

 

[imath]E = m c^2[/imath], and [imath]m = m_0 \left( 1 - \left( \frac{v}{c} \right)^2 \right)^{-\frac12}[/imath]

 

Do the algebra, taking care with the units, you get

 

[math]\frac{v}{c} \, \dot= \, \sqrt{1 -\left( \frac{510998.91}{542998.91} \right)^2} \, \dot= \, 0.338217354[/math]

 

which seems about right

 

Up the voltage to a particle accelerator-like 1000000000 V, you get

 

[math]\frac{v}{c} \, \dot= \, \sqrt{1 -\left( \frac{510998.91}{1000510998.91} \right)^2} \, \dot= \, 0.999999869573378[/math]

 

which also seems about right.

[kjf]

Wow!

Thanks for the reply - Don't worry, I'm not going anywhere - I'll post a thread about myself soon enough!

 

Just a quick question, though...

It's a potential difference of 32,000V, how does that translate to 32,000 eV?

Is the "1.6 x 10^-19 Coulomb per electron" a way of converting to joules, rather than eV?

If so, I've been wrong with most of my calculations! (I don't have an answers booklet to check, I just knew this one was wrong because of the "faster than c" velocity haha).

[CraigD]

It's a potential difference of 32,000V, how does that translate to 32,000 eV?

The electron volt is a unit of energy defined as

equal to the amount of kinetic energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt.

Put another way,

[imath]1 \,\mbox{volt} = \frac{1 \,\mbox{joule}}{1 \,\mbox{coulomb}}[/imath]

so

[math]1 \,\mbox{eV} = 1 \,\mbox{volt} \cdot Q_{\mbox{electron}} = \frac{1 \,\mbox{joule} \cdot Q_{\mbox{electron}}}{1 \,\mbox{coulomb}} \dot= 1.6021765 \times 10^{-19} \,\mbox{joule}[/math]

Is the "1.6 x 10^-19 Coulomb per electron" a way of converting to joules, rather than eV?

Yes.

 

You could do all of the above calculation in Joules, foot-pounds force, BTUs, or any other unit of energy, and get the same result for [imath]\frac{v}{c}[/imath]. eVs are just a convenient unit to use when you’re working with volt units and electrons, or other charged subatomic particles, as the avoid inexact conversion factors and large negative powers of ten.