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The \mathrm{K} \alpha \mathrm{x}-ray emission line of tungsten occurs at \mathrm{\lambda=0.021 \mathrm{~nm}}. The energy difference between \mathrm{\mathrm{K}\: and \: \mathrm{L}} levels in this atoms is about :
 

Option: 1

0.51 \mathrm{MeV}


 


Option: 2

1.2 \mathrm{MeV}
 


Option: 3

59 \mathrm{keV}
 


Option: 4

13.6 \mathrm{eV}


Answers (1)

best_answer

\mathrm{\lambda_{k_\alpha}=0.021 \mathrm{~nm}=0.21 \AA}

since, \mathrm{\lambda_k \propto} corresponds to the transition of an electron from L-shell to K-shell, therefore

\mathrm{\mathrm{E}_{\mathrm{L}}-\mathrm{E}_{\mathrm{K}} =(\text { in eV })=\frac{12375}{\lambda(\text { in } \AA)}=\frac{12375}{0.21} }

\mathrm{ \approx 58928 \mathrm{eV} }

\mathrm{ \Delta \mathrm{E}(\text { in } \mathrm{eV})= \frac{12375}{1(\text { in } \AA)} \text { comes for } \mathrm{E}=\frac{h c}{\lambda} }




 

Posted by

Riya

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