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(B) The Bragg's law for diffraction is given by $2d \sin 	heta = n\lambda$,where $	heta$ is the glancing angle. From the geometry of the problem,the

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$50$

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(B) The Bragg's law for diffraction is given by $2d \sin 	heta = n\lambda$,where $	heta$ is the glancing angle. From the geometry of the problem,the angle of incidence $i$ is measured with respect to the normal to the crystal plane. Thus,the glancing angle $	heta = 90^\circ - i = 90^\circ - 30^\circ = 60^\circ$. Assuming first-order diffraction $(n = 1)$,we have $2d \sin(60^\circ) = \lambda$. Substituting $d = 1 \ \mathring{A}$,we get $\lambda = 2 	imes 1 	imes \frac{\sqrt{3}}{2} = \sqrt{3} \ \mathring{A} \approx 1.732 \ \mathring{A}$. The de Broglie wavelength of an electron accelerated by potential $V$ is given by $\lambda = \sqrt{\frac{150}{V}} \ \mathring{A}$. Equating the two: $\sqrt{3} = \sqrt{\frac{150}{V}}$. Squaring both sides: $3 = \frac{150}{V}$. Therefore,$V = \frac{150}{3} = 50 \ V$.

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