\(2 d \sin \theta= n \lambda\)

angle of incidence \(=90- i\)

Also \(n=1\) for \(1^{\text {st }}\) order diffraction.

Therefore,

\(2 \times 1 \times 10^{-10} \sin (90-30)=1 \times \lambda\)

\(\Rightarrow \lambda=2 \times 10^{-10} \frac{\sqrt{3}}{2}=\sqrt{3} \times 10^{-10} m\)

We have de-Broglie wavelength, \(\lambda=\frac{ h }{ mv }\)

and also \(\frac{1}{2} mv ^{2}= eV \Rightarrow v =\sqrt{\frac{2 eV }{ m }}\)

Substituting in de-Broglie wavength, we get,

\(\lambda=\frac{ h }{\sqrt{2 meV }}=\sqrt{3} \times 10^{-10} m\)

\(V=49.86\;V\)