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BHU PMT BHU PMT (Mains) Solved Paper-2006

  • question_answer
    The\[{{K}_{a}}\]X-ray emission line of tungsten occurs at\[\lambda =0.021\text{ }mm\]. The energy difference between K and L levels in this atom is about:

    A)  0.51 MeV           

    B)  1.2 MeV

    C)  59 keV                                

    D)  13.6 eV

    Correct Answer: C

    Solution :

                     \[^{\lambda }{{k}_{\alpha }}=0.021\,nm=0.21{\AA}\] Since\[^{\lambda }{{k}_{\alpha }}\]corresponds to the transition of an electron form L-shell to K-shell, therefore \[{{E}_{L}}-{{E}_{k}}=(in\,eV)=\frac{12375}{\lambda (in\,{\AA})}=\frac{12375}{0.21}\]                                 \[\cong 58928\,eV\] Or                           \[\Delta E\approx 59\,keV\] Note:\[\Delta E(in\text{ }eV)=\frac{12375}{\lambda (in\,{\AA})}\]comes for \[E=\frac{hc}{\lambda }.\]


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