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Crystal diffraction experiments can be p...

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? An X-ray photon or the electron? (For quantitative comparison, take the wavelength of the probe equal to `1Å`, which is of the order of interatomic spacing in the lattice), `m_(e)=9.11xx10^(-31)kg`.

A

X-ray

B

Electron

C

both have same energy

D

cannot be determined by crystal diffraction.

Text Solution

Verified by Experts

The correct Answer is:
A
For an accelerated electron beam , the de - Broglie matter wave equation states that
`lamda=h/(sqrt(2emV)) = h/(sqrt(2mK)) implies K =(h^2)/(2mlamda^2)`
For X - rays , photos of same wavelength `lamda =1 Å`
`E.=hupsilon=(hc)/lamda" or " K/(E.) = ((h^2)/(2mlamda^2))/((hc)/lamda)`
or `K/(E.) = h^2 /(2mlamda^(2) )xxlamda/(hc) =h/(2mclamda)`
`:. K/(E.) = h/(2mclamda)`
where h `= 6.6 xx 10^(-34) Js, m =9.1 xx 10^(-31) kg`
`c = 3 xx 10^(8) m//s lamda = 1 Å= 10^(-10)m`
` K/(E.) = ((6.6 xx10^(-34)))/(2 xx 9.1 xx 10^(-31)xx3 xx 10^(8)xx 10^(-10)) = 11/91`
`implies K lt E.`
`implies K lt E.`
`:.` Energy possessed by X - ray is more than electron.
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