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K_(alpha) radiation of Mo (Z = 42) has a wavelength of 0.71Å. Calculate wavelength of the corresponding radiation of Cu (Z=29) |
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Answer» `1.45Å` `V=a(Z-SIGMA)^(2)` for `K_(alpha)` radiationscreening CONSTANT `sigma =1 and v=(c)/(lambda)` `:. (1)/(lambda) prop (Z-1)^(2)` `(lambda_(Cu))/(lambda_(MO))=((Z_(Mo)-1)^(2))/((Z_(Cu)-1)^(1))=(41^(2))/(28^(2))` `:. lambda_(Mo)=0*71Å` Then `lambda_(Cu)=(0.71)Åxx(41^(2))/(28^(2))=1*52Å` |
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