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. Thankskuruman said:Therefore an electron that moves through a potential difference of 75 keV acquires kinetic energy of ... ?
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 said: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
Thanks, so by my understand of your reply, 1 eV = (1.6*10-19) (1 V)kuruman said: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?
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!axer said:V=75,000/1.6*10-19
No! Electron volts is a unit of energy, not a unit of potential difference! Units!axer said:1 eV = (1.6*10-19) (1 V)
Ok i see.. so 1 eV= 1.6*10-19 JOrodruin said:No! Electron volts is a unit of energy, not a unit of potential difference! Units!
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?axer said:Ok i see.. so 1 eV= 1.6*10-19 J
axer said:V=W/Q
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.axer said: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!
Potential difference in an X-ray tube is the difference in electrical potential between the cathode and anode, which creates an electric field to accelerate electrons towards the anode. It is also known as voltage or electrostatic potential.
Potential difference can be calculated using the formula V = I x R, where V is the potential difference, I is the current, and R is the resistance. In an X-ray tube, the potential difference is determined by the voltage applied to the tube.
Potential difference is crucial in an X-ray tube as it determines the speed and energy of the electrons, which in turn affects the quality and intensity of the X-ray beam produced. It also plays a role in controlling the exposure time and dose received by the patient.
The main factors that can affect potential difference in an X-ray tube include the voltage applied, the distance between the cathode and anode, and the type of material used for the anode. Additionally, any changes in the electrical circuit, such as a decrease in resistance, can also affect potential difference.
Potential difference can be adjusted by changing the voltage applied to the X-ray tube. This can be done manually by the operator or automatically through the use of exposure control systems. The voltage can typically range from 50-150 kV for diagnostic X-ray tubes.