All questions
AP Physics 2
15.5 The Photoelectric Effect
15.3 Emission and Absorption Spectra
AdvancedMCQMathematicalConceptual15k
An energy level diagram showing three horizontal parallel lines representing stationary energy states of hydrogen. The lowest horizontal line is labeled n = 1 at the left and E_1 = -13.6 eV at the right. The middle horizontal line is labeled n = 2 at the left and E_2 = -3.4 eV at the right. The top horizontal line is labeled n = 4 at the left and E_4 = -0.85 eV at the right. On the left side, under the header Sample 1, a single downward vertical arrow extends directly from n = 4 to n = 1. On the right side, under the header Sample 2, two connected downward vertical arrows extend in sequence: the first from n = 4 to n = 2, and the second from n = 2 to n = 1. No other labels, lines, text, or axes appear.
Energy level transitions for Sample 1 and Sample 2.
Two identical gas samples containing hydrogen atoms in the excited state \(n = 4\) undergo de-excitation to the ground state \(n = 1\). In Sample 1, all atoms decay via a single direct transition (\(n = 4 \rightarrow n = 1\)). In Sample 2, all atoms decay via a two-step cascade transition (\(n = 4 \rightarrow n = 2 \rightarrow n = 1\)). The photons emitted from each sample strike identical metal plates, each having a work function \(\Phi = 4.0\text{ eV}\). How do the total energy of all photons emitted (\(E_{\text{total}}\)) and the maximum kinetic energy of ejected photoelectrons (\(K_{\text{max}}\)) compare between Sample 1 and Sample 2?

Log In to Continue

Accounts are free! Log in to try this question, see explanations, save progress, and more!

Tools for a 5