The electron in a hydrogen atom falls from an excited energy level to the ground state in two steps, causing the emission of photons with wavelengths of 2624 nm and 97.2 nm, respectively. What is the principal quantum number or shell of the initial excited energy level from which the electron falls
A) 2
B) 3
C) 4
D) 6
E) 8

Question

Quizplus · Accepted Answer

We can use the Rydberg formula to determine the initial shell number. The formula is:

1/λ = R (1/n1^2 - 1/n2^2)

where λ is the wavelength of the emitted photon, R is the Rydberg constant (1.097 × 10^7 m^-1), n1 is the initial shell number, and n2 is the final shell number (which is 1 for the ground state).

For the first emission with a wavelength of 2624 nm, we can plug in the values and solve for n1:

1/2624 nm = 1.097 × 10^7 m^-1 (1/n1^2 - 1/1^2)

n1 = 4

So the electron started in the fourth energy level.

For the second emission with a wavelength of 97.2 nm, we can repeat the calculation:

1/97.2 nm = 1.097 × 10^7 m^-1 (1/4^2 - 1/1^2)

n1 = 6

Therefore, the electron fell from the sixth energy level to the ground state in two steps.

The Electron in a Hydrogen Atom Falls from an Excited