A) [tex]5.0\cdot 10^{-11} m[/tex]
The energy of an x-ray photon used for single dental x-rays is
[tex]E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J[/tex]
The energy of a photon is related to its wavelength by the equation
[tex]E=\frac{hc}{\lambda}[/tex]
where
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant
[tex]c=3\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda[/tex] is the wavelength
Re-arranging the equation for the wavelength, we find
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m[/tex]
B) [tex]2.0\cdot 10^{-11} m[/tex]
The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is
[tex]E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J[/tex]
And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m[/tex]
