1. **State the problem:** We have an electron transitioning from energy level A (ground state) to energy level C (excited state) in an atom. We need to find the wavelength of the photon absorbed or emitted during this transition. 2. **Given data:** - Energy at ground state (A): $150$ zJ = $150 \times 10^{-21}$ J - Energy at excited state (C): $1200$ zJ = $1200 \times 10^{-21}$ J 3. **Determine if photon is absorbed or emitted:** Since the electron moves from a lower energy level (A) to a higher energy level (C), it absorbs energy. So, a photon is absorbed. 4. **Calculate the energy difference $\Delta E$ of the photon:** $$\Delta E = E_C - E_A = 1200 \times 10^{-21} - 150 \times 10^{-21} = 1050 \times 10^{-21} = 1.05 \times 10^{-18} \text{ J}$$ 5. **Use the energy-wavelength relation:** $$E = \frac{hc}{\lambda}$$ where $h = 6.626 \times 10^{-34}$ J\cdots (Planck's constant), $c = 3.00 \times 10^{8}$ m/s (speed of light), and $\lambda$ is the wavelength in meters. 6. **Solve for wavelength $\lambda$:** $$\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{1.05 \times 10^{-18}}$$ 7. **Calculate $\lambda$:** $$\lambda = \frac{1.9878 \times 10^{-25}}{1.05 \times 10^{-18}} = 1.893 \times 10^{-7} \text{ m}$$ 8. **Convert meters to nanometers:** $$1.893 \times 10^{-7} \text{ m} = 189.3 \text{ nm}$$ 9. **Round to 3 significant digits:** $$\boxed{189 \text{ nm}}$$ **Final answer:** The wavelength of the absorbed photon is 189 nm.