For hydrogen,what is the wavelength of the photon emitted when an electron drops from a 4d orbital to a 2p orbital in a hydrogen atom? A)656.3 nm B)486.2 nm C)364.6 nm D)2.057 × 10^-3 nm

Question

Quizplus · Accepted Answer

The equation for calculating the wavelength of a photon emitted during a transition of an electron in a hydrogen atom is:
ΔE = hc/λ 
where ΔE is the energy difference between the initial and final states, h is Planck's constant, c is the speed of light and λ is the wavelength of the emitted photon.
For the given transition from 4d to 2p, the energy difference can be calculated as:
ΔE = E(2p) - E(4d) = (-0.544 eV) - (-0.136 eV) = -0.408 eV
Using the value of h and c:
λ = hc/ΔE = (6.626 x 10^-34 J.s x 3.0 x 10^8 m/s) / (-0.408 eV x 1.602 x 10^-19 J/eV) = 486.2 nm
Therefore, the correct choice is B.

For Hydrogen,what Is the Wavelength of the Photon Emitted When