Question

The Ka X-ray emission line of tungsten occurs at I = 0.021 nm. The energy difference between K and L levels in this atoms is about: (IIT JEE 1997)

• A. 0.51 MeV
• B. 1.2 MeV
• C. 59 keV
• D. 13.6 eV

Answer 1

The correct option is C

59 keV

We know that, the energy difference is carried by the photon. So, the energy of the mentioned photon will give us the desired result.
$E=\frac{hc}{\lambda }=\frac{1240eV-nm}{0.021}=59,047eV$
Which is approximately equal to 59KeV