1. **State the problem:** We are given energy levels of an electron in zeptojoules (zJ) at states A (ground state), B (first excited state), and C (second excited state). The electron transitions from state A to state C, and we need to determine if a photon is absorbed or emitted and calculate the wavelength of that photon. 2. **Given data:** - Energy at ground state (A): $E_A = 150$ zJ - Energy at first excited state (B): $E_B = 350$ zJ - Energy at second excited state (C): $E_C = 500$ zJ 3. **Determine if photon is absorbed or emitted:** Since the electron moves from a lower energy state (A) to a higher energy state (C), it must absorb energy. Therefore, a photon is **absorbed**. 4. **Calculate the energy difference of the photon:** $$\Delta E = E_C - E_A = 500 - 150 = 350 \text{ zJ}$$ 5. **Convert energy from zeptojoules to joules:** $$350 \text{ zJ} = 350 \times 10^{-21} \text{ J} = 3.5 \times 10^{-19} \text{ J}$$ 6. **Use the photon energy-wavelength relation:** $$E = \frac{hc}{\lambda}$$ where - $h = 6.626 \times 10^{-34}$ Js (Planck's constant) - $c = 3.00 \times 10^{8}$ m/s (speed of light) - $\lambda$ is the wavelength in meters Rearranged to solve for wavelength: $$\lambda = \frac{hc}{E}$$ 7. **Calculate wavelength:** $$\lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{3.5 \times 10^{-19}} = \frac{1.9878 \times 10^{-25}}{3.5 \times 10^{-19}}$$ 8. **Simplify:** $$\lambda = 5.68 \times 10^{-7} \text{ meters}$$ 9. **Convert meters to nanometers:** $$5.68 \times 10^{-7} \text{ m} = 568 \text{ nm}$$ 10. **Final answer:** The photon absorbed has a wavelength of approximately **568 nm**.