Find out the atomic number of the element which gives X-ray of minimum wavelength 0.252 nm of K series `(R=1.09737xx10^(7)m^(-1))`

Options

A. 2
B. 200
C. 20
D. 2000

Answer

C

Explanation

`(1)/(lambda)=R(Z-1)^(2)[(1)/(1)^(2)-(1)/(n^(2))]`
For the minimum wavelength we have to take `n=oo`
`:. (1)/(lambda_(min))=R(Z-1)^(2)[1-0]`
`:.(1)/(lambda_(min))=R(Z-1)^(2)`
`:.[Z-1]^(2)=(1)/(R lambda_(min))=(1)/(1.097xx10^(7)xx252xx10^(-12))`
`:.[Z-1]^(2)=(10^(5))/(1.097xx252)=0.0036174xx10^(5)`
`:.[Z-1]^(2)=361.74`
`Z-1=19.02=19`
`:.Z=19+1=20`