For hydrogen,what is the wavelength of the photon emitted when an electron drops from a 4p orbital to a 2s orbital in a hydrogen atom? The Rydberg constant is 1.097 × 10^-2 nm^-1. A) 656.3 nm B) 486.2 nm C) 364.6 nm D) 2.057 × 10^-3 nm

Question

Quizplus · Accepted Answer

The energy difference between the 4p and 2s orbitals can be calculated using the Rydberg formula:

1/λ = R*(1/n2^2 - 1/n1^2)

where λ is the wavelength of the photon emitted, R is the Rydberg constant, and n1 and n2 are the initial and final energy levels of the electron, respectively.

Plugging in the values n1=4 and n2=2, and the given value of R, we get:

1/λ = (1.097 x 10^7 m^-1) * (1/2^2 - 1/4^2)
1/λ = 10970000 m^-1 * (3/16)
1/λ = 2057.142857 m^-1

Converting this to nanometers, we get:

λ = 1/2057.142857 x 10^9 m/nm
λ ≈ 486.2 nm

Therefore, the correct answer is B) 486.2 nm.

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