1. Single dental x-rays:
![5.0\cdot 10^(-11)m](https://img.qammunity.org/2020/formulas/physics/high-school/ndclvmci3uy3879br7gvt8gnwb5bwiv47z.png)
The energy of the photon is
![E=25 keV = 25,000 eV](https://img.qammunity.org/2020/formulas/physics/high-school/4fzncwhgo308j98joilrokikhmbcw30bzk.png)
Using the conversion factor
![1 eV=1.6\cdot 10^(-19) J](https://img.qammunity.org/2020/formulas/physics/high-school/ympbw887qawl4eyookss32zazljuzgc2p4.png)
we can convert it into Joules:
![E=(25,000 eV)(1.6\cdot 10^(-19)J/eV)=4\cdot 10^(-15) J](https://img.qammunity.org/2020/formulas/physics/high-school/6vyo4dlxiixndanuo0cso2c4v5qjj7yfcu.png)
The relationship between photon energy and wavelength is
![\lambda=(hc)/(E)](https://img.qammunity.org/2020/formulas/physics/high-school/xvfldutqd45p1j55qd9gdy092m3sgqlpye.png)
where
is the Planck constant
is the speed of light
E is the energy
Substituting into the formula, we find
![\lambda=((6.63\cdot 10^(-34)Js)(3\cdot 10^8 m/s))/(4\cdot 10^(-15) J)=5.0\cdot 10^(-11)m](https://img.qammunity.org/2020/formulas/physics/high-school/molohqylm0pd5donoeuln1a2lvu4p8fw4o.png)
2. Microtomography:
![2.0\cdot 10^(-11) m](https://img.qammunity.org/2020/formulas/physics/high-school/my2c66al465iuatyycv2dmzbar7hgakfu2.png)
The energy of these photons is 2.5 times greater, so
![E=(2.5)(4\cdot 10^(-15) J)=1\cdot 10^(-14) J](https://img.qammunity.org/2020/formulas/physics/high-school/30nhoyc0us8rwlez2ar7xwf0w1qtik5t95.png)
And by applying the same formula used at point 1, we find the corresponding wavelength:
![\lambda=((6.63\cdot 10^(-34)Js)(3\cdot 10^8 m/s))/(1\cdot 10^(-14) J)=2.0\cdot 10^(-11)m](https://img.qammunity.org/2020/formulas/physics/high-school/o8hd4dh5fsajzp2twevwk01wu4pbnb01fm.png)