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. `0.50`
D. `0.48`

Answer

B

Explanation

`sqrt(C/lambda)=a (Z-b)`
`b=1 sqrt(lambda_(Cu)/lambda_(Me))=((Z_(Me)-1)/(Z_(Cu)-1))`
`lambda_(Cu)/lambda_(Me)=(41/28)^(2)=2.14`