`K_(alpha)` radiation of Mo (Z = 42) has a wavelength of `0.71Å`. Calculate wavelength of the corresponding radiation of Cu (Z=29)

Options

A. `1.45Å`
B. `1.52Å`
C. `0.52Å`
D. `1.02Å`

Answer

B

Explanation

From the Mosley.s law, we have
`v=a(Z-sigma)^(2)`
for `K_(alpha)` radiation screening 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`