In making a transition from state $n = 1$ to state $n = 2$, the hydrogen atom must
A) emit a photon of energy $10.2 \mathrm { eV }$.
B) absorb a photon of energy $13.58 \mathrm { eV }$.
C) emit a photon of energy $13.58 \mathrm { eV }$.
D) absorb a photon of energy $10.2 \mathrm { eV }$.

Question

Quizplus · Accepted Answer

For a hydrogen atom to transition from a lower energy state ($n = 1$) to a higher energy state ($n = 2$), it must absorb energy. The energy difference between these two states corresponds to a photon of energy $10.2 \mathrm{eV}$.

In Making a Transition from State n=1n = 1n=1 To State n=2n = 2n=2

In Making a Transition from State $n = 1$ To State $n = 2$