What is the minimum potential difference?

[sweetipie2216]

What is the minimum potential difference applied to an x-ray tube if x-rays of wavelength 0.250 nm are produced?



E=hf
f=c/wavelength

f=3e8/.250e-9
=1.2e18

then i applied it to the first equation using Plancks constant as h and my final answer came out to be 7.95e-16 KV.

the answer is wrong. can you please help me out? thanks.

 

[Chewy0087]

half way there,

the work done accelerating the electrons is equal to $$eV$$, but you can also find out the energy of the xrays, as you've shown, these energies should be equal.

you worked out the energy of the xrays, not the voltage.

 

[kerimek]

The minimum potential difference required for the production of x-rays of a specific wavelength is known as the "cut-off voltage" or the "threshold voltage." This is the minimum amount of energy needed to overcome the binding energy of electrons in the target material and produce x-rays of that particular wavelength. It can be calculated using the equation E = hf = hc/wavelength, where h is Planck's constant, c is the speed of light, and wavelength is the desired x-ray wavelength.

In this case, if we use a wavelength of 0.250 nm, we can calculate the minimum potential difference as follows:

E = (6.626 x 10^-34 J.s)(3 x 10^8 m/s)/(0.250 x 10^-9 m)
= 7.95 x 10^-16 J

To convert this to kilovolts, we can use the equation V = E/q, where V is the voltage, E is the energy, and q is the charge. Since we know that 1 electron has a charge of 1.602 x 10^-19 C, we can calculate the minimum potential difference as:

V = (7.95 x 10^-16 J)/(1.602 x 10^-19 C)
= 495.6 kV

Therefore, the minimum potential difference applied to an x-ray tube to produce x-rays of wavelength 0.250 nm is 495.6 kV. It is important to note that this is the minimum value and the actual potential difference used in an x-ray tube may be higher to account for inefficiencies and other factors.