Question

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
A. `1.99`
B. `2.14`
C. `0.50`
D. `0.48`

Answer 1

Correct Answer - B
`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`