When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the photon emitted is:
A) 0
B) 1.5 eV
C) 9.1 eV
D) 12.1 eV
E) 13.6 eV

Question

Quizplus · Accepted Answer

The formula for calculating the energy of a photon emitted during a transition is:
ΔE = E_final - E_initial = hv 
Where ΔE is the change in energy, E_final is the energy of the final state, E_initial is the energy of the initial state, h is Planck's constant, and v is the frequency of the emitted photon.

For the given transition, the initial state is the second excited state, which has an energy of -3.4 eV. The final state is the ground state, which has an energy of -13.6 eV. Substituting these values into the formula:
ΔE = (-13.6 eV) - (-3.4 eV) = -10.2 eV
The negative sign indicates that energy is being emitted.

Now, using the formula E = hv, we can solve for the frequency of the emitted photon:
v = ΔE/h = (-10.2 eV)/(6.626 x 10^-34 J s) = -1.54 x 10^15 Hz
Again, the negative sign indicates that the photon is being emitted (rather than absorbed).

Finally, we can use the formula E = hv to convert the frequency to energy:
E = hv = (-1.54 x 10^15 Hz)(6.626 x 10^-34 J s) = -1.02 x 10^-18 J = -6.36 eV
Again, the negative sign indicates that energy is being emitted.

Thus, the energy of the photon emitted during this transition is approximately 6.36 eV. The closest choice is D) 12.1 eV, which is the energy difference between the first excited state (-3.4 eV) and the ground state (-13.6 eV), rather than the energy difference between the second excited state and the ground state. So the answer is D.

When a Hydrogen Atom Makes the Transition from the Second