Potential Difference for X-Ray Emission

[steven10137]

[SOLVED] Potential Difference argh ...

Homework Statement

What is the minimum potential difference that must be applied across an X-ray tube to observe a line of frequency 1.61x10^16Hz

Homework Equations

OK well we can define potential difference as the amount of energy required to emit this X-ray yeah?

The Attempt at a Solution

<br /> \begin{array}{l}<br /> E = hf \\ <br /> = \left( {6.63 \times 10^{ - 34} } \right)\left( {1.61 \times 10^{16} } \right) \\ <br /> = 1.07 \times 10^{ - 17} J \\ <br /> \end{array}<br />

But I thought that potential difference was measured in volts?

Just need clarification here
Thanks
Steven

 

[Shooting Star]

The x-ray emission in an x-ray tube is generally brake radiation. This is emitted by electrons when they are stopped at the anode. The electrons are accelerated due to potential difference after they are emitted by the cathode until they reach the anode. There, the KE of an electron is converted into the energy of a photon. For simplicity, let’s assume that all the KE of an e- is converted into one x-ray photon.

Then E=hf, is the energy of the photon, which is equal to the KE of the e- after it was accelerated across the tube. Work done on e- = V*e, where e is the magnitude of the electronic charge, and V is the potential difference.

Then, V*e=hf => V=hf/e should give you the minimum potential difference.

Again, this is only if the process of emission is that of brake radiation, which is generally what happens in x-ray tubes.

 

[steven10137]

thanks!
makes perfect sense