A sodium atom is in one of the states labelled 'Lowest excited levels'. It remains in that state for an average time of `10^(-8)` sec. before it makes a transition back to a ground state. What is the uncertianty in enegry of that excited state ?

Options

A. `6.56 xx 10^(-8) eV`
B. `2 xx 10^(-8) eV`
C. `10^(-8) eV`
D. `8 xx 10^(-8) eV`

Answer

A

Explanation

The average time that the atom spends in this excited state is equal to `Deltat`, so by using `DeltaE. Deltat = (h)/(2pi)`
`rArr` Uncertainty in energy `= (h//2pi)/(Deltat)`
`= (6.6 xx 10^(-34))/(2 xx 3.14 xx 10^(-8)) = 1.05 xx 10^(-26)J = 6.56 xx 10^(-8) eV`