Final answer:
To find the frequency of light emitted by an electron moving from orbit n=2 to n=1 inside of a He+ ion, the Bohr model and Planck's equation are used to calculate the energy difference between orbits and thereby find the photon's frequency.
Step-by-step explanation:
Calculating the Frequency of Light in a Helium Ion Transition
The question involves calculating the frequency of light emitted when an electron in a helium ion (He+) transitions from a higher energy orbit to a lower one. Specifically, from n1=2 to n2=1. Using the Bohr model of the atom, this can be understood as the electron dropping from one quantized energy level to another, releasing a photon whose energy corresponds to the difference between these two energy states.
The energy difference (ΔE) between two orbits in a hydrogen-like atom is given by the formula:
ΔE= Z2RH(1/n12 - 1/n22)
Here, Z is the atomic number of helium (Z=2), RH is the Rydberg constant for hydrogen (approximately 13.6 eV), n1 and n2 are the principal quantum numbers of the initial and final orbits, respectively. The frequency (f) of the emitted photon is related to the energy difference by Planck's equation:
E= hf, where h is Planck's constant
Now, rearranging for f and using the values for a helium ion, we get:
f = ΔE/h = Z2RH(1/n12 - 1/n22)/h
By substituting the known values into this equation, we can calculate the frequency of the photon produced when the electron falls from orbit 2 to orbit 1 in a helium ion.