If `lambda_(Cu)` is the wavelength of `K_(alpha)` X-ray line of copper (atomic number `29`) and `lambda_(Mo)` is the wavelength of the `K_(alpha)`X-ray line of molybdenum (atomic number `42`),then the ratio `lambda_(Cu)//lambda_(Mo)` is close to

Options

A. 1.99
B. 2.14
C. 0.5
D. 0.48

Answer

B

Explanation

`(lambda_(Cu))/(lambda_(Mo))=((Z_(Mo)-1)/(Z_(Cu)-1))^(2)=(41xx41)/(28xx28)=(1681)/(784)=2.144`