Q. The$\ K_{a}\ X$-ray emission line of tungsten occurs at $\lambda = 0.021nm.$ 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) $13.6\ eV$
Explanation
(c) $E_{K} - E_{L} = \frac{hc}{\lambda} = \frac{\left( 6.6 \times 10^{- 34} \right)\left( 3 \times 10^{8} \right)}{\left( 0.021 \times 10^{- 9} \right)\left( 1.6 \times 10^{- 19} \right)}eV = 59KeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.