Final answer:
The wavelength of the photon emitted by the hydrogen atom when its electron drops from the n=4 to n=2 state is approximately 64.63 nm.
Step-by-step explanation:
To calculate the wavelength of a photon emitted when an electron drops from the n=4 to n=2 state in a hydrogen atom, we can use the formula:
wavelength = hc/ΔE
Where h is Planck's constant (6.626 x 10-34 J·s), c is the speed of light (3 x 108 m/s), and ΔE is the change in the energy levels. In this case, ΔE = En=2 - En=4. Using the formula, we can calculate the wavelength to be:
wavelength = (6.626 x 10-34 J·s)(3 x 108 m/s) / (En=4 - En=2)
Using the given information, we can substitute the energy levels of hydrogen in the formula and solve for the wavelength. The final answer will be in meters, so to convert it to nanometers, we need to multiply by 109.
Therefore, the wavelength of the photon emitted by the hydrogen atom is approximately 64.63 nm.