Answer:
a. TRUE
b. FALSE
c. TRUE
Explanation:
a. A hydrogen atom in the n=3 state can emit light at only two specific wavelengths. TRUE. The emission of light is produced by the reduction of the state, in this case, of the atom. If the atom is in state 3 the reduction of state will occurs just to state 1 and to state 2. Each reduction of state will produce one specific wavelenght
b. A hydrogen atom in the n=2 state is at a lower energy than one in the n=1 state. FALSE. The n=1 state is the basal state where the atom has the lower energy.
c. The energy of an emitted photon equals the energy difference of the two states involved in the emission. TRUE. The energy of an emitted photon is calculated with the energy of the high state minus the energy of the low state.
I hope it helps!
Answer:
a. A hydrogen atom in the n=3 state can emit light at only two specific wavelengths.
c. The energy of an emitted photon equals the energy difference of the two states involved in the emission.
Explanation:
When a hydrogen atom absorbs a quantum of energy, it promotes its electron into a higher energy state (meaning an orbit of greater radius to Bohr). There are only specific energy states (levels) possible, and we number them n=1, n=2, n=3 etc.
When this electron descends to a lower energy, the atom emits a photon of light. The Law of Conservation of Energy requires that the energy of the photon equal the difference in energy between the initial and final states. Hence (c) is true.
The electron can drop into any lower state, and if it is initially at n=3, there are only two states lower, and only two energy changes are possible. Hence (a) is true.
However, for an electron at n=2, there is a lower state (n=1), so (b) is false.
Only a and c are true.
