The wavelength of `K_(a)` , X-rays for lead isotopes `Pb^(208),Pb^(206),Pb^(204)" are"lambda_(1),lambda_(2)" and"lambda_(3),` resperctively. Then

Options

A. `lambda_(2)=sqrt(lambda_(1)lambda_(3))`
B. `lambda_(2)=lambda_(1)+lambda_(3)`
C. `lambda_(2)=lambda_(1)lambda_(3)`
D. `lambda_(2)=(lambda_(1))/(lambda_(3))`

Answer

A

Explanation

Wavelengths of the `K_(alpha)` lines for given isotopes of lead (Pb) can be given by a general expression
`(1)/(lambda)=R(Z-1)^(2)((1)/(1^(2))-(1)/(2^(2)))`
where R = Rydberg's constant, Z = atomic number of the isotopes. Though `Pb^(208), Pb^(206) and Pb^(204)` have different atomic masses, Z will be same for them i.e. 82.
`therefore (1)/(lambda_(1))=R(8201)^(2)((1)/(1^(2))-(1)/(2^(2)))=(3)/(4)R(81)^(2)`
`(1)/(lambda_(2))=(3)/(4)R(81^(2)) and (1)/(lambda_(3))=(3)/(4)R(81^(2))`
`rArr ((1)/(lambda_(2)))^(2)=(1)/(lambda_(1))xx(1)/(lambda_(3))rArr lambda_(2) = sqrt(lambda_(1)lambda_(3))`