## University Physics with Modern Physics (14th Edition)

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.