The distance between crystals plane is approximately $10^{-10}m$ what approximate photon energy is required to observe Bragg reflection from the planes. Can someone explain how the answer was calculated.
The answer is $4*10^{3}eV$ (electron volts).
Can someone please give a good detailed answer, i thought i was genius at Bragg's law until i came across this.
Best Answer
Here is my approach - which doesn't quite give the same answer you were looking for...
From Bragg's Law, we know
$$n\lambda = 2d\sin\theta$$
This tells us that the smaller the angle $\theta$, the shorter the wavelength - and the shorter the wavelength, the greater the energy, since wavelength and energy are related through the relationship:
$$E = \frac{hc}{\lambda}$$
The maximum allowed wavelength (lowest energy) is the one for which $n=1$ and $\theta = 90°$. Then $\lambda = 2d$.
The rest is substitution of numbers in the above equations. A handy shortcut is given on wikipedia:
$$E = \frac{1.2398}{\lambda (µm)} ~\rm{eV}$$
with the wavelength given in µm. Given $\lambda = 2d = 2\times10^{-10}$ m, the above evaluates to
$$E = \frac{1.24}{2\times 10^{-4}~µm} = 6.2 ~\rm{keV}$$
Not sure why that is not exactly the same as the answer you were given. There seems to be about a $\sqrt{2}$ difference...