Electrons in a TV tube accelerated from rest.

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SUMMARY

Electrons in a TV tube are accelerated from rest through a 25-kV potential difference, resulting in a kinetic energy of 4 x 10^-16 J per electron. The correct calculation for the speed of the electrons upon hitting the screen is derived using the formula K = (1/2)mv^2, leading to a final speed of 9.37 x 10^7 m/s. A common error in the calculations was using an incorrect voltage value of 2.5 x 10^3 J/C instead of the correct 25 x 10^3 J/C. This mistake resulted in an initial incorrect speed calculation of 2.9 x 10^7 m/s.

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  • Understanding of kinetic energy (K = (1/2)mv^2)
  • Familiarity with electric potential (voltage) and its units (J/C)
  • Basic knowledge of electron charge (1.6 x 10^-19 C)
  • Newtonian mechanics principles
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  • Learn about the relationship between voltage, energy, and charge
  • Study the implications of unit conversions in physics calculations
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_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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