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An electron beam can undergo diffraction...

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength is equal to 1.6 A?

Text Solution

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For an electron
`1/2"mu"^(2)=eV`
where V is accelerating potential
`lamda=h/("mu")`
`:.1/2m(h/(m lamda))^(2)=eV`
`:.V=1/2xx(h^(2))/(m lamda^(2)e)=(1xx(6.625xx10^(-34))^(2))/(2xx9.108xx10^(-31)xx(1.54xx10^(-10))^(2)xx1.602x10^(-19))=63.3` volt
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