What is the frequency of a photon resulting from the transition of
n=6 -->n=1?
for what atom. H? U? Na?
If for H, use one of the next two or three posts to solve.
The frequency of a photon resulting from the transition of an electron from the n=6 to n=1 energy state depends on the specific atom in question. In this case, we have the options of hydrogen (H), uranium (U), or sodium (Na).
To find the frequency of the photon for hydrogen (H), we can use the formula:
ΔE = hf
where ΔE represents the change in energy, h is the Planck's constant (6.626 × 10^-34 J·s), and f is the frequency of the photon.
The change in energy (ΔE) can be calculated using the formula:
ΔE = E_final - E_initial
where E_final corresponds to the energy of the final state (n=1) and E_initial corresponds to the energy of the initial state (n=6).
For hydrogen, the energy levels are given by the formula:
E_n = -13.6 eV / n^2
where E_n represents the energy level for a given value of n.
Plugging in the initial and final energy levels for hydrogen (n=6 and n=1) and using the proper unit conversion from electron volts (eV) to joules (J), we can calculate ΔE. Then, using the value of h, we can find the frequency (f) of the photon.
The calculations will be shown in the next few posts.