Electron falling between two orbits produces certain wavelenght of light

[Nano]

Homework Statement

Between which two orbits of the Bohr hydrogen atom must an electron fall to produce light of wavelength 954.8 nm ?

Homework Equations

E = hv
E = R(1/n1^2 - 1/n2^2)

The Attempt at a Solution

I've got it down to the differences in the reciprocals of the squares of the energy levels = .09548. I don't know how to calculate each n though. I've tried randomly plugging in different n's to see if i get .09548, but that's not getting me anywhere.

 

[nate808]

Any help would be great!


Hi there,

To calculate the two orbits between which the electron must fall, we can use the Rydberg formula:

1/λ = R(1/n1^2 - 1/n2^2)

Where λ is the wavelength of the emitted light, R is the Rydberg constant (1.097 x 10^7 m^-1), and n1 and n2 are the initial and final energy levels of the electron, respectively.

Solving for n1 and n2, we get:

n1 = sqrt(R/(1/λ + 1/n2^2))

n2 = sqrt(R/(1/λ - 1/n1^2))

Plugging in the given wavelength of 954.8 nm, we get:

n1 = sqrt(1.097 x 10^7 m^-1/(1/954.8 x 10^-9 m + 1/n2^2))

n2 = sqrt(1.097 x 10^7 m^-1/(1/954.8 x 10^-9 m - 1/n1^2))

To find the two orbits, we can try different values of n1 and n2 and see which combination gives us a difference of 0.09548. For example, if we let n1 = 2 and n2 = 3, we get a difference of 0.09548, meaning the electron must fall from the n=3 to n=2 orbit to produce light of wavelength 954.8 nm.

I hope this helps! Let me know if you have any further questions.