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  • The  x-Ray emission line of tungsten occurs at <img alt="\mathrm{ \lambda=0.0

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

Option: 1

0.51 ~\mathrm{MeV}


Option: 2

59~ \mathrm{keV}


Option: 3

1.2 ~\mathrm{MeV}


Option: 4

13.6 ~\mathrm{eV}


Answers (1)

best_answer

Given, \mathrm{ \lambda_{k \alpha}=0.021 ~\mathrm{nm}=0.21 \hat{A}}
since, \mathrm{ \lambda_{k \alpha}} corresponding to the transition of an electron from L- shell to k-shell therefore,

\mathrm{ E_L-E_k=(\text { inc })=\frac{12375}{\lambda(\operatorname{in} A)}=\frac{12375}{0.21}}

\begin{aligned} \mathrm{E_L-E_K} & =58928 \mathrm{eV} \\ \Delta \mathrm{E }& =59 \mathrm{keV} \end{aligned}

Posted by

Devendra Khairwa

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