If the 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`.

Options

A. `1.52 Å`
B. `2.52 Å`
C. `0.52 Å`
D. `4.52Å`

Answer

A

Explanation

`((Z_(MO)-1)^2)/( (Z_(Cu ) -1)^2 ) = ( lamda _(cu ) )/( lamda _(MO ) ) or ((41 )/( 28 ))^2 = ( lamda _(cu )/( 0.71 ))`
` therefore lamda_(cu ) = 0.71 xx ((41)/( 28) ) ^2 = 1.52 Å`