Final answer:
To find the initial energy level when an electron drops from a higher energy level to n=2 and emits a photon with a wavelength of 410nm,
Step-by-step explanation:
When an electron drops from a higher energy level to n=2 and emits a photon with a wavelength of 410nm, we can use the energy equation to find the initial energy level.
The energy of a photon is given by the equation E = (hc)/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength. Rearranging the equation, we have E = (hc)/λ, and solving for E, we get E = (6.63 x 10^-34 J·s)(3 x 10^8 m/s)/(410 x 10^-9 m) = 4.84 x 10^-19 J.
This energy corresponds to the difference in energy levels between the initial and final states of the electron. To determine the initial energy level, we need to find the difference in energy levels between n=2 and the initial state.
We use the equation ΔE = E_initial - E_final, where ΔE is the difference in energy levels, E_initial is the energy of the initial level, and E_final is the energy of the final level.
Rearranging the equation, we get E_initial = ΔE + E_final. Since we know the final energy is 4.84 x 10^-19 J and n=2, we can plug in the values to find the initial energy level.