Answer :

To find the minimum x-ray wavelength produced for a dental x-ray machine operated at a voltage of 37.6 kV, you can use the formula relating the energy of the photons to their wavelength:

[tex]\text{Energy} (E) = \frac{hc}{\lambda}[/tex]

Where:

  • [tex]h[/tex] is Planck's constant [tex](6.626 \times 10^{-34} \text{ J} \cdot \text{s})[/tex].
  • [tex]c[/tex] is the speed of light [tex](3.00 \times 10^8 \text{ m/s})[/tex].
  • [tex]\lambda[/tex] is the wavelength in meters.

The energy of the x-rays can also be expressed in electron volts (eV), using the given voltage:

[tex]E = eV[/tex]

where [tex]e[/tex] is the charge of an electron [tex](1.602 \times 10^{-19} \text{ C})[/tex]. For a potential difference of 37.6 kV, you first convert this to volts:

[tex]E = 37.6 \times 10^3 \text{ V} \times 1.602 \times 10^{-19} \text{ J/eV}[/tex]

This gives us the energy in joules. Now, equating these two expressions for energy and solving for [tex]\lambda[/tex]:

[tex]\frac{hc}{\lambda} = 37.6 \times 10^3 \times 1.602 \times 10^{-19}[/tex]

Rearranging for [tex]\lambda[/tex]:

[tex]\lambda = \frac{hc}{37.6 \times 10^3 \times 1.602 \times 10^{-19}}[/tex]

Substitute the values for [tex]h[/tex] and [tex]c[/tex]:

[tex]\lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{37.6 \times 10^3 \times 1.602 \times 10^{-19}}[/tex]

Calculating this gives:

[tex]\lambda \approx 3.30 \times 10^{-11} \text{ meters}[/tex]

Since we need the wavelength in nanometers (1 nm = [tex]10^{-9}[/tex] meters):

[tex]\lambda \approx 0.0330 \text{ nm}[/tex]

Therefore, the minimum x-ray wavelength produced is approximately 0.0330 nm.

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Rewritten by : Barada