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AIIMS 2009
Exam
AIIMS
Year
2009
Subject
Physics
Chapter
Atomic Physics

AIIMS 2009 · Physics · Atomic Physics

The wavelengths of K_α X -rays for lead isotopes Pb^ 208 , ~Pb^ 206...

The wavelengths of KαX\mathrm{K}_\alpha \mathrm{X}-rays for lead isotopes Pb208, Pb206\mathrm{Pb}^{208}, \mathrm{~Pb}^{206} and Pb204\mathrm{Pb}^{204} are λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3 respectively. Then

Options

  1. Aλ2=λ1λ3\lambda_2=\sqrt{\lambda_1 \lambda_3}
  2. Bλ2=λ1+λ3\lambda_2=\lambda_1+\lambda_3
  3. Cλ2=λ1λ3\lambda_2=\lambda_1 \lambda_3
  4. Dλ2=λ1λ3\lambda_2=\frac{\lambda_1}{\lambda_3}

Correct answer

A. λ2=λ1λ3\lambda_2=\sqrt{\lambda_1 \lambda_3}

Step-by-step solution

Wavelengths of the KαK_\alpha lines for given isotopes of lead (Pb)(\mathrm{Pb}) can be given by a general expression 1λ=R(Z−1)2\frac{1}{\lambda}=R(Z-1)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)wherewhereR=Rydberg′sconstant,Rydberg's constant,Z=atomicnumberoftheisotopes.Thoughatomic number of the isotopes. Though\mathrm{Pb}^{208}, \mathrm{~Pb}^{206} and Pb204\mathrm{Pb}^{204} have different atomic masses, ZZ will be same for them i.e. 82 . ∴1λ1=R(82−1)2(112−122)=34R(81)21λ2=34R(812) and 1λ3=34R(812)⇒(1λ2)2=1λ1×1λ3⇒λ2=λ1λ3\begin{aligned} & \therefore \frac{1}{\lambda_1}=R(82-1)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R(81)^2 \\ & \frac{1}{\lambda_2}=\frac{3}{4} R\left(81^2\right) \text{ and } \frac{1}{\lambda_3}=\frac{3}{4} R\left(81^2\right) \\ & \Rightarrow\left(\frac{1}{\lambda_2}\right)^2=\frac{1}{\lambda_1} \times \frac{1}{\lambda_3} \Rightarrow \lambda_2=\sqrt{\lambda_1 \lambda_3}\end{aligned}
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