if `lambda_(Cu)` is the wavelength of `K_alpha`, X-ray line fo 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
(a) 1.99
(b) 2.14
(c ) 0.50
(d) 0.48

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`