Calculating Potential difference (X Ray tube)

[axer]

Homework Statement

If the electrons hit the tungsten target with an energy of 75 keV, find the potential difference.

Homework Equations

The Attempt at a Solution

I tried this: (W=eV)
V=W/Q
V=eV/Q
V=75,000/1.6*10^-19
V= 4.6*10²³ V,

This seems wrong because all examples in my textbook show V is in the thousands but not this big..

 

[kuruman]

An electron that moves through a potential difference of 1 Volt acquires kinetic energy of 1 eV. Therefore an electron that moves through a potential difference of 75 keV acquires kinetic energy of ... ?

 

[axer]

Therefore an electron that moves through a potential difference of 75 keV acquires kinetic energy of ... ?

Thanks for your quick reply, in the question the potential difference was not given and must be calculated. IIm just confused why you said pd of 75keV. Thanks

 

[kuruman]

Thanks for your quick reply, in the question the potential difference was not given and must be calculated. IIm just confused why you said pd of 75keV. Thanks

I wrote pd of 75 keV not 75 eV. Let me restate this. The kinetic energy of an electron starting from rest and moving through a pd of 1 V is 1 eV. In other words, the electron trades potential energy for kinetic energy as it moves from a region of high potential energy to a region of low potential energy. What if the acquired linetic energy is 75 keV (75,000 eV) instead of just 1 eV?

 

[axer]

I wrote pd of 75 keV not 75 eV. Let me restate this. The kinetic energy of an electron starting from rest and moving through a pd of 1 V is 1 eV. In other words, the electron trades potential energy for kinetic energy as it moves from a region of high potential energy to a region of low potential energy. What if the acquired linetic energy is 75 keV (75,000 eV) instead of just 1 eV?

Thanks, so by my understand of your reply, 1 eV = (1.6*10^-19) (1 V)

So if we had 75,000 eV, then the potential difference is also 75,000.
Answer: 75,000 V.

Right?

 

[Orodruin]

V=75,000/1.6*10-19

To rsetate what is being hintef at, this would be the potential difference (in volts) if the energy was 75 kJ, not 75 keV. 1 J is not 1 eV. Units are important, never neglect them and always write them out!

 

[Orodruin]

1 eV = (1.6*10-19) (1 V)

No! Electron volts is a unit of energy, not a unit of potential difference! Units!

 

[axer]

No! Electron volts is a unit of energy, not a unit of potential difference! Units!

Ok i see.. so 1 eV= 1.6*10^-19 J

75 keV = 1.6*10^-19 (75000)
= 1.2*10^-14 J.

this is the energy in Joule of an electron with an energy of 75 keV.

If right.. How do I use it to find V?

If I am being wrong/confusing, would you please show me the answer so I can find where I am thinking wrong. Thanks!

 

[kuruman]

So you have found that 75 keV is 1.2×10^-14J. Use energy conservation and the equation W = eV that you quoted in post #1 to find V. The result may be surprising but not a coincidence.

 

[Orodruin]

Ok i see.. so 1 eV= 1.6*10-19 J

Right, now go back to the relation that expresses the energy gained in terms of the charge and the potential difference. What does it tell you?

V=W/Q

 

[axer]

so V=1.2*10^-14/1.6*10^-19
V= 75.000 V
Oh, wait is it actually 75,000... this apply to all questions when given eV and asked to find V.

Thanks both!

 

[Orodruin]

so V=1.2*10^-14/1.6*10^-19
V= 75.000 V
Oh, wait is it actually 75,000... this apply to all questions when given eV and asked to find V.

Thanks both!

Yes, but again, always write out the units. Note that you had the right answer in #5, but you got it in a way that was not completely waterproof. Always write out your units, not only will it let you check that your answer has the correct units, but it is particularly useful to get your argument correct.