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If the Kalpha radiation of Mo (Z = 42) h...

If the `K_alpha` radiation of `Mo (Z = 42)` has a wevelength of `0.71Å`, calculate wevelength of the corresponding radiation of `Cu,i.e. , k_(alpha)for Cu (Z = 29)` assuming b = 1.

Text Solution

Verified by Experts

From Moseley.s law, we have ` v = K(Z-b)^(2)`
For `K_(a)` radiation, Screening constant `b = 1 and v = (c/lambda)`
`therefore 1/lambda propto (Z-1)^(2)`
`therefore` `lambdacu/lambdaMo = (Z_(Mo - 1)^(2))/(Z_(cu) - 1)^(2))`
=` (41^(2))/(28^(2))`
Now given `lambda_(Mo) = 0.71 A`
`therefore lambda _(Cu) = (0.71 A) xx 41^(2)/28^(2)`
= 1.52 A
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