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(B) For diffraction to occur in the nickel granules,the condition is $2d \sin 	heta = n \lambda$. Since $\sin 	heta \le 1$,the maximum wavelength th

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less than $\frac{h^2}{8md^2}$

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(B) For diffraction to occur in the nickel granules,the condition is $2d \sin 	heta = n \lambda$. Since $\sin 	heta \le 1$,the maximum wavelength that can be diffracted is $\lambda_{\max} = 2d$ (for $n=1$). Electrons with wavelengths $\lambda > 2d$ cannot be diffracted by the nickel granules and will pass through the pipe. Using the de Broglie relation $\lambda = \frac{h}{p}$,the condition $\lambda > 2d$ implies $\frac{h}{p} > 2d$,or $p The kinetic energy $E$ of these electrons is $E = \frac{p^2}{2m}$. Substituting the inequality for $p$,we get $E Thus,the electrons that pass through the pipe have energy less than $\frac{h^2}{8md^2}$.

$A$ long pipe is filled with small granules of nickel. The interatomic spacing for nickel is $d$. $A$ beam of electrons having different energies is incident on the left flat face of the pipe. Mass of electron is $m$. The beam of electrons coming out of the right flat face has energy:

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