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Find out the difference in energy between \mathrm{K \, \&\, L} levels in \mathrm{Mo} if it was observed that \mathrm{K_\alpha\: X-ray } emission line of this atom is formed at \mathrm{\lambda=0.031 \mathrm{~nm} }.

Option: 1

0. 40 \, \mathrm{keV}


Option: 2

40\, \mathrm{keV}


Option: 3

40 \, \mathrm{MeV}


Option: 4

13.6\, \mathrm{eV}


Answers (1)

best_answer

\lambda_{K_\alpha} =0.031 \mathrm{~nm}
           =0.31 \mathrm{~A}^{\circ}

This \lambda corresponds to the transition of an electron from L shell to K shell.
E =\frac{h c}{\lambda}
E_L-E_k =\frac{12375}{\lambda\left(\text { in } A^{\circ}\right)} \mathrm{eV}
                    =\frac{12375}{0.31} \mathrm{eV}
                    =39919.35 \, \mathrm{eV}
\Delta E=39.9\, \, \mathrm{KeV }\sim 40\,\, KeV

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Nehul

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