1.

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 ...

Answer»

`lambda_(2)= sqrt(lambda_(1)lambda_(3))`
`lambda_(3)=lambda_(1)+lambda_(3)`
`lambda_(2)=lambda_(1)lambda_(3)`
`lambda_(2)=(lambda_(1))/(lambda_(3))`

Solution :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))`


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