The wavelength of `K_(alpha)` X-rays for lead isotopes `P_(b)^(208),P_(b)^(206) and P_(b)^(204)` are `lambda_(1),lambda_(2), and lambda_(3)`, respectively. Then ...

Options

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

Answer

A

Explanation

Wavelength 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 `P_(b)^(208) P_(b)^(206), P_(b)^(204)` have different atomic masses, Z will be same for them i.e 82.
`:.(1)/(lambda_(1))=R(82-1)^(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))+(1)/(lambda_(3)) rArr lambda_(2)= sqrt(lambda_(1)lambda_(3))`