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  • The wave length of the characteristic <img alt="\mathrm{ \mathrm{X}-ray \mathrm{K}_\alpha}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cmathrm%7BX%7D-ray%20%5Cmathrm

The wave length of the characteristic \mathrm{ \mathrm{X}-ray \mathrm{K}_\alpha} line emitted by a hydrogen like atom is \mathrm{ 0.32 \AA}. The wave length of \mathrm{\mathrm{K}_\beta} line emitted by the same element is:




 

Option: 1

\mathrm{0.18 \AA}
 


Option: 2

\mathrm{\quad 0.48 \AA}
 


Option: 3

\mathrm{0.27 \AA}


Option: 4

\mathrm{\quad 0.38 \AA}


Answers (1)

best_answer

The wavelength of \mathrm{x}-rays emitted for \mathrm{K}_\alpha \: \text{line and}\: \mathrm{K}_\betaline are given by \mathrm{\lambda_{\mathrm{a}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{x}}-\mathrm{E}_{\mathrm{L}}}\: and \: \: \lambda_\beta=\frac{h c}{E_K-E_M} }

\mathrm{\therefore \frac{\lambda_\alpha}{\lambda_\beta}=\frac{E_K-E_M}{E_K-E_L}=\frac{\frac{1}{1}-\frac{1}{3^2}}{\frac{1}{1}-\frac{1}{2^2}}=\frac{32}{27} }

\mathrm{\therefore \lambda_\beta=\frac{27}{32} \times \lambda_\alpha=\frac{27}{32} \times 0.32=0.27 \AA }

 

Posted by

Suraj Bhandari

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