Between which two orbits of the Bohr hydrogen atom must an electron fall to produce light of wavelength 434.2 ?
Well, if an electron falls between orbits, it's probably because it tripped over its own feet. Those electrons, always so clumsy! But let me get serious for a moment. The difference in energy between two Bohr hydrogen atom orbits corresponds to the wavelength of light emitted when an electron makes that jump. So, to find the answer to your question, we'll use the formula for the difference in energy levels of the hydrogen atom. Assuming you're referring to the Balmer series, we can use the formula:
1/λ = R_H * (1/n_final^2 - 1/n_initial^2),
where λ is the wavelength, R_H is the Rydberg constant, and n_final and n_initial are the final and initial principal quantum numbers, respectively. Rearranging the formula gives us:
1/n_final^2 - 1/n_initial^2 = λ/R_H.
Now we can plug in the given wavelength of 434.2 nm and the known values for the Rydberg constant and solve the equation. Using a little bit of algebraic magic, we find that the electron falls between the 4th and 2nd orbits. Don't worry, the electron is just enjoying a little rollercoaster ride inside the hydrogen atom! Wheeeee!
But remember, my friend, this is just a simplified model. The actual behavior of electrons gets a lot more complicated and unpredictable. So, take this answer with a grain of salt, or maybe a dash of lemon pepper if you're into that kind of thing.