Final answer:
The wavelength of a photon needed to excite an electron from E1 to E4 in a hypothetical atom is 141 nm.
Step-by-step explanation:
The student's question revolves around the concept of atomic excitations and de-excitations and the resulting emission or absorption of photons, which is rooted in quantum mechanics and is an essential part of Physics.
To calculate the wavelength of a photon needed to excite an electron from energy level E1 to E4, we first determine the energy difference between these levels:
ΔE = E4 - E1
ΔE = (-1.0 x 10^-19 J) - (-15 x 10^-19 J)
ΔE = 14 x 10^-19 J
We then use the equation relating energy (E) and wavelength (λ) of a photon:
E = h * c / λ
where h is Planck's constant (6.626 x 10^-34 J*s) and c is the speed of light in a vacuum (3.0 x 10^8 m/s).
Now we can solve for λ:
λ = h * c / ΔE
λ = (6.626 x 10^-34 J*s) * (3.0 x 10^8 m/s) / (14 x 10^-19 J)
λ = 1.41 x 10^-7 meters or 141 nm