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

For the emission of X-ray line of wavelength `lambda`,
`1/lambda=R(Z-1)^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]`
`1/(lambda_(Cu))=R(Z_(Cu)-1)^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]........(i)`
`1/(lambda_(Mo))=R(Z_(Mo)-1)^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]........(ii)`
Dividing (ii) by (i), we get
`(lambda_(Cu))/(lambda_(Mo))=((Z_(Mo)-1)^(2))/((Z_(Cu)-1)^(2)) = ((92-1)^(2))/((29-1)^(2))=(41/28)^(2) = 2.14`