An element with Z = 11 emits `Kalpha`-X-ray with wavelength `lamda`. The atomic no. of element which emits `Kalpha` X-ray of wavelength `4lamda` is:

Options

A. 6
B. 4
C. 11
D. 44

Answer

A

Explanation

`sqrt(V)=a(z-b)`
`v=a^(2)(z-b)^(2)`
`:. (c)/(lambda)=a^(2)(z-b)^(2)`
`:.(lambda_(1))/(lambda_(2))=((z_(2)-1)^(2))/((z_(1)-z)^(2))` here b=1
put `z_(1)=11, lambda_(1)= lambda, lambda_(2)= 4lambda_(2)` & find `z_(2)`
Then `z_(2)=6.`