We have,
2dsinθ=nλ
angle of incidence = 90−i
Given i=30o, d=1×10−10m, h=6.6×10−34, m=9.1×10−31 and e=1.6×10−19
Also n=1 for 1st order diffraction.
Therefore,
2×1×10−10sin(90−30)=1×λ
⟹λ=2×10−10√32=√3×10−10m
We have de-Broglie wavelength, λ=hmv
and also 12mv2=eV ⟹v=√2eVm
Substituting in de-Broglie wavength, we get,
λ=h√2meV= √3×10−10m
⟹V=49.86V