The wavelength of `K_(alpha)` X-rays of two metals `A and B` are` 4 // 1875 R and 1// 675 R`, respectively , where `R` is rydberg 's constant. The number of electron lying between `A and B` according to this lineis

Options

A. 3
B. 6
C. 5
D. 4

Answer

D

Explanation

`1/lambda = R(Z-1)^(2)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
For wavelength of `K_(alpha), n_(1) = 1` to `n_(2) = 2`
`(1875R)/(4) = R(Z_(A) - 1)^(2) [1-1/4] = "…"(i)`
and `(675R)/(1) = R[Z_(B) - 1]^(2) [1-1/4]"……"(ii)`
By solving eq. (i) & (ii) we get
`Z_(A) = 26` and `Z_(B) = 31`
[Four elements lie between these two]