A hydrogen atom is excited to the n = 10 level.Its decay to the n = 4 level is detected in a photographic plate.What is the frequency of the light photographed? A)1.73 × 10¹⁴ Hz B)1740 Hz C)1740 nm D)1.46 × 10^-9km

Question

Quizplus · Accepted Answer

The frequency of the light emitted during the decay is given by the formula:
f = (E_n1 - E_n2)/h
where E_n1 and E_n2 are the energies of the initial and final states, respectively, and h is Planck's constant.
For this problem, we have:
E_n1 = -13.6 / 10^2 = -0.136 eV
E_n2 = -13.6 / 4^2 = -0.85 eV
Converting these energies to joules:
E_n1 = -2.18 x 10^-18 J
E_n2 = -1.36 x 10^-18 J
Substituting into the formula:
f = (1.36 - (-2.18)) x 10^18 / 6.626 x 10^-34
f = 1.73 x 10^15 Hz
Which corresponds to choice A) 1.73 × 10^15 Hz.

A Hydrogen Atom Is Excited to the N = 10