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characteristic-spectra of x-ray is shown in figure. Which is \mathrm{k_\alpha } and \mathrm{k_\beta }-

Option: 1

\mathrm{k_\alpha=k_\beta=0}


Option: 2

\mathrm{k_\alpha=k_\beta=1}


Option: 3

\mathrm{k_\alpha=2, k_\beta =1}


Option: 4

\mathrm{k_\alpha=k_\beta=-1}


Answers (1)

best_answer

Here, \mathrm{\lambda=\frac{h c}{\Delta E}}
Since energy difference of \mathrm{ k_\alpha} is less than \mathrm{ k_\beta}

\begin{aligned} \mathrm{\Delta E_{K_\alpha}} &\mathrm{<\Delta E_{K_\beta}} \\ \Rightarrow \quad\mathrm{ \lambda_{k_\beta} }&\mathrm{ <\lambda_{k \alpha} }\end{aligned}
Therefore 1 is \mathrm{k_\beta} and 2 is \mathrm{k_\alpha}

Posted by

Ritika Jonwal

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