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The wavelength of the charastistic \mathrm{x}-ray \mathrm{k_\alpha} line emitted by a hydrogen-like element is \mathrm{0.32 \AA}. Cal culate the wavelength of \mathrm{K_\beta} line emitted by the same element.

Option: 1

\mathrm{0.27\ \AA }


Option: 2

\mathrm{0.46\ \AA }


Option: 3

\mathrm{0.75\ \AA }


Option: 4

\mathrm{0.98\ \AA }


Answers (1)

best_answer

For hydrogen like atom - 
\mathrm{\frac{1}{\lambda}=z^2 R\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]}
Case \mathrm{I} - for \mathrm{k\alpha -} line 
\mathrm{\begin{aligned} & \frac{1}{\lambda k_\alpha}=z^2 R\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ & \frac{1}{\lambda k_\alpha}=\frac{3 z^2 R}{4}----\left ( 1 \right ) \end{aligned}}
Case \mathrm{II} - for \mathrm{k\ \beta\ -} line 
\mathrm{\frac{1}{\lambda k_\beta}=2^2 R\left[\frac{1}{1^2}-\frac{1}{3^2}\right]}
\mathrm{\frac{1}{\lambda_{k_\beta}}=\frac{8 z^2 R}{g}----\text { (2) }}
Form equation \mathrm{\text { (1) }} and \mathrm{\text { (2) }}, we have 
\mathrm{\frac{\lambda_{k_\beta}}{\lambda_{k_\alpha}}=\frac{3 / 4}{8 / 9}=\frac{3 \times 9}{4 \times 8}}
\mathrm{\begin{aligned} & \frac{\lambda_{k \beta}}{\lambda k_\alpha}=\frac{27}{32} \\ & \lambda_{k_\beta}=\frac{27}{32} \lambda_{k_\alpha}=\frac{27}{32} \times 0.32 \AA \end{aligned}}
\mathrm{\lambda_{k_\beta}=0.27 \AA}

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vinayak

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