Electrons in a TV tube accelerated from rest.

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Homework Help Overview

The problem involves electrons being accelerated from rest through a 25-kV potential difference in a TV tube, with the goal of determining their speed upon impact with the screen. The subject area encompasses concepts from electromagnetism and kinematics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between voltage, energy, and charge, and how to calculate kinetic energy from the potential difference. There are attempts to apply Newtonian mechanics and electromagnetic principles to derive the speed of the electrons.

Discussion Status

Some participants have offered hints regarding the relationship between volts and energy, while others have pointed out potential errors in calculations. There is an ongoing exploration of unit consistency and the implications of using incorrect values in calculations.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of guidance provided. There is a noted discrepancy between calculated results and textbook answers, prompting further investigation into the calculations and assumptions made.

_F_
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Homework Statement


Electrons in a TV tube are accelerated from rest through a 25-kV potential difference. With what speed do they hit the TV?

V = 25*10^3 J/C


Homework Equations


I'm not really sure here. I'm thinking you have to use Newtonian mechanics somewhere here.
So,
x(t) = (1/2)(a_x)t^2 + (v_0)t +x_0
(v_f)^2 = (v_0)^2 + 2(a_x)d

You also need electomagnetic equations.
So,
delta-V = delta-U/q = -E*delta-r

The Attempt at a Solution


This is where I have trouble. I don't even know where to start. Any pointers in the right direction (that aren't too vague!) would greatly be appreciated.
 
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Welcome to PF!

_F_ said:
Electrons in a TV tube are accelerated from rest through a 25-kV potential difference. With what speed do they hit the TV?

V = 25*10^3 J/C

Hi _F_! Welcome to PF! :smile:

Hint: volts = energy/charge (as you can see from the J/C units).

So what KE does each electron have after going through a 25kV potential difference? :smile:
 


tiny-tim said:
Hi _F_! Welcome to PF! :smile:

Hint: volts = energy/charge (as you can see from the J/C units).

So what KE does each electron have after going through a 25kV potential difference? :smile:

Okay, so K = (1/2)mv^2

We have volts in J/C, so to get a value of just energy would we multiply V*q?
If so:
V*q = (2.5*10^3 J/C)(1.6*10^-19 C) = 4*10^-16 J.

So K = 4*10^-16 J,
Then
K/[(1/2)m] = v^2 => (4*10^-16 J)/[(1/2)(9.1*10^-31 kg) = 8.79*10^14 J/kg

So v^2 = 8.79*10^14 J/kg
Then, v = sqrt(8.79*10^14 J/kg) = 2.9*10^7 m/s

But this is not what the back of my book says. In the back, the answer is 9.37*10^7 m/s.
 
_F_ said:
Okay, so K = (1/2)mv^2

We have volts in J/C, so to get a value of just energy would we multiply V*q?
If so:
V*q = (2.5*10^3 J/C)(1.6*10^-19 C) = 4*10^-16 J.

So K = 4*10^-16 J,
Then
K/[(1/2)m] = v^2 => (4*10^-16 J)/[(1/2)(9.1*10^-31 kg) = 8.79*10^14 J/kg

So v^2 = 8.79*10^14 J/kg
Then, v = sqrt(8.79*10^14 J/kg) = 2.9*10^7 m/s

But this is not what the back of my book says. In the back, the answer is 9.37*10^7 m/s.

Hi _F_! :smile:

Hint: check your units … you're out by a factor of √10. :wink:
 
tiny-tim said:
Hi _F_! :smile:

Hint: check your units … you're out by a factor of √10. :wink:

J = kg*m^2/s^2

V = [J/C]
Vq = [J]
K = [J]
v^2 = 2K/m = [m^2/s^2]
v = [m/s]

Where did I go wrong?
 
_F_ said:
J = kg*m^2/s^2

V = [J/C]
Vq = [J]
K = [J]
v^2 = 2K/m = [m^2/s^2]
v = [m/s]

Where did I go wrong?

V is given to be 25*10^3 J/C. You used 2.5*10^3.
 
Dick said:
V is given to be 25*10^3 J/C. You used 2.5*10^3.

Ack! Thanks. Such a stupid mistake. :rolleyes:
 

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