3. When an electron in a hydrogen atom drops from the 5 th to the 2 nd energy level, \( 4.58 \times 10^{-18} \mathrm{~J} \) of energy is released. Find the frequency of the photon that is produced? \( \left(6.91 \times 10^{-14} \mathrm{~s}^{-1}\right) \)

To find the frequency of the photon produced when the electron transitions from the 5th to the 2nd energy level in a hydrogen atom, we can use the energy-frequency relation given by \( E = h \cdot f \), where \( E \) is the energy in joules, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), and \( f \) is the frequency in hertz. By rearranging the equation to solve for frequency, we have: \[ f = \frac{E}{h} \] Substituting in the given value: \[ f = \frac{4.58 \times 10^{-18} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} \] Calculating this gives approximately \( f \approx 6.91 \times 10^{15} \, \text{s}^{-1} \), which is in the range of visible light, making it quite fascinating that such transitions produce light we can actually see! A neat piece of information is that these energy transitions in atoms are responsible for the colors we perceive in fireworks or neon signs, where different elements emit distinct frequencies of light! When an electron drops energy levels, it emits specific wavelengths, creating those vibrant effects.