The `X`- ray wavelength of `L_(alpha)` line of platinum `(Z = 78)` is `1.30 Å`. The `X` - ray wavelength of `L_(alpha)` line of Molybdenum `(Z = 42)` is

Options

A. `5.41 Å`
B. `4.20 Å`
C. `2.70 Å`
D. `1.35 Å`

Answer

A

Explanation

The wavelength of `L_(alpha)` line is given by
`(1)/(lambda) = R( z - 7.4)^(2) ((1)/(2^(2)) - (1)/(3^(2))) rArr lambda prop (1)/((Z - 7.4)^(2))`
`rArr (lambda_(1))/(lambda_(2)) = ((z_(2) - 7.4)^(2))/((z_(1) - 7.4)^(2))`
`rArr (1.30)/(lambda_(2)) = ((42 - 7.4)^(2))/((78 - 7.4)^(2)) rArr lambda_(2) = 5.41 Å`
( `Delta E = ` Energy radiated when e - jumps from , higher energy orbit to lower energy orbit)
`because (Delta E)_(k_(beta)) gt (Delta E)_(k_(alpha)) gt (Delta E)_(L_(alpha)) :. lambda'_(alpha) gt lambda_(alpha) gt lambda_(beta)`
Also `(Delta E)_(k_(beta)) = (Delta E)_(k_(alpha)) + (Delta E)_(L_(alpha))`
`rArr (hc)/(lambda_(beta)) = (hc)/(lambda_(alpha)) + (hc )/(lambda'_(alpha)) rArr (1)/(lambda_(beta)) = (1)/(lambda_(alpha)) + (1)/(lambda'_(alpha))`