Q. A sodium atom is in one of the states labeled ‘Lowest excited levels’. It remains in that state for an average time of $10^{- 8}s$, before it makes a transition back to a ground state. What is the uncertainty in energy of that excited state
A) $6.56 \times 10^{- 8}eV$
B) $2 \times 10^{- 8}eV$
C) $10^{- 8}eV$
D) $8 \times 10^{- 8}eV$
Explanation
(a) The average time that the atom spends in this excited sate is equal to $\mathrm{\Delta}t$, so by using $\mathrm{\Delta}E.\mathrm{\Delta}t = \frac{h}{2\pi}$
$\Rightarrow$ Unertainty in energy $= \frac{h/2\pi}{\mathrm{\Delta}t}$
$= \frac{6.6 \times 10^{- 34}}{2 \times 3.14 \times 10^{- 8}} = 1.05 \times 10^{- 26}J = 6.56 \times 10^{- 8}eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.