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An electron beam can undergo diffraction by crystals.Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54A^{\large\circ}$?

$(a)\;62.3v\qquad(b)\;63.3v\qquad(c)\;61.5v\qquad(d)\;60v$

Can you answer this question?
 
 

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For an electron $\large\frac{1}{2}$$mu^2=e.V$
and $\lambda=\large\frac{h}{mu}$
Thus $\large\frac{1}{2}m$$\large\frac{h^2}{m^2\lambda^2}$$=e.V$
$V=\large\frac{1\times (6.62\times 10^{-34})^2}{2\times 9.108\times 10^{-31}\times (1.54\times 10^{-10})^2\times 1.602\times 10^{-19}}$
$\Rightarrow 63.3Volt$
Hence (b) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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