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Kalpha wavelength emitted by an atom of ...

`K_alpha` wavelength emitted by an atom of atomic number Z=11 is `lambda`. Find the atomic number for an atom that emits `K_alpha` radiation with wavelength `4lambda`.
(a) Z=6 (b) Z=4
(c ) Z=11 (d) Z=44.

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Verified by Experts

The correct Answer is:
6
`1/lambda prop (z-1)^(2)`
`therefore lambda_(1)/lambda_(2) =(z_(2)-z_(1))^(2)/(z_(1)-1)^(2)`
`1/4 = ((z_(2)-1)/(11-1))^(2)`
Solving this, we get, `Z_(2) = 6`
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