Answer
See explanation.
Work Step by Step
a. The photon energy equals the transition energy, 3.50 eV.
$$\lambda =\frac{hc}{E}=\frac{1.24\times10^{-6}eV\cdot m}{3.50eV}=3.54\times10^{-7}m$$
b. The Heisenberg uncertainty principle limits our knowledge.
$$\Delta E \Delta t \geq \frac{\hbar}{2}$$
Find the minimum uncertainty in energy.
$$\Delta E = \frac{\hbar}{2\;\Delta t}$$
$$\Delta E = 2.6\times10^{-29}J=1.6\times10^{-10}eV $$
The uncertainty in the energy could be larger than that quantity, but it could not be smaller.
