The decay of excited states in atoms and nuclei often leave the system in another, albeit lower-energy, excited state. (a) One example is the decay between two excited states of the nucleus of Ti. The upper state has a lifetime of 1.4 ps, the lower state 3.0 ps. What is the frac tional uncertainty AE/E in the energy of 1.3117-MeV gamma rays connecting the two states?

To solve this problem, we need to use the Heisenberg Uncertainty Principle, which states that the uncertainty in energy (ΔE) and the uncertainty in time (Δt) are related by the equation:

ΔE * Δt ≥ h/4π

where h is Planck's constant.

Given that the lifetime of the upper state is 1.4 ps and the lower state is 3.0 ps, we can take the average of these two times as the uncertainty in time (Δt).

So, Δt = (1.4 ps + 3.0 ps) / 2 = 2.2 ps = 2.2 * 10^-12 s

Now, we can solve for ΔE using the Heisenberg Uncertainty Principle:

ΔE ≥ h / (4π * Δt)

Substituting the values, we get:

ΔE ≥ (6.626 * 10^-34 J.s) / (4π * 2.2 * 10^-12 s)

ΔE ≥ 2.39 * 10^-23 J

But the energy of the gamma rays is given in MeV, so we need to convert this energy to Joules.

1 MeV = 1.602 * 10^-13 J

So, E = 1.3117 MeV = 1.3117 * 1.602 * 10^-13 J = 2.10 * 10^-13 J

Finally, we can find the fractional uncertainty ΔE/E:

ΔE/E = (2.39 * 10^-23 J) / (2.10 * 10^-13 J) = 1.14 * 10^-10

So, the fractional uncertainty in the energy of the gamma rays is 1.14 * 10^-10.