Practicing Success
We wish to observe an object which is 2.5 Å in size. The minimum energy photon that can be used - |
5 KeV 8 KeV 10 KeV 12 KeV |
5 KeV |
In order for scattering to occur, the wavelength of the waves must be of the same order of magnitude or smaller than the size of the object being observed. Hence the largest possible wavelength we can use in the present problem is $λ_{max} = 2.5 Å$. Hence minimum energy is $E_{min}=hv_{min}=\frac{hc}{λ_{max}}=\frac{12.40×10^3}{2.5Å}eV.Å=4.96×10^3eV=5KeV$ |