Of the following transitions in the Bohr hydrogen atom, the ____ transition results in the emission of the highest-energy photon.

a.n=1, n=6
b.n=6, n=1
c.n=6,n=3
d.n=3,n=6
e.n=1,n=4

B: n=6, n=1 is highest-energy photon. Emission is going from a higher energy level down to a lower energy level, so that will rule out a,d,e. So between b and c, it will be b because if you study the n=1 energy level you will see that it is the ground state energy level, which means that energy is the greatest in this energy level.

The transition that results in the emission of the highest-energy photon in the Bohr hydrogen atom is the transition from a higher energy state to a lower energy state.

By comparing the given transitions, we can see that option b) n=6 to n=1 represents a transition from a higher energy state (n=6) to a lower energy state (n=1). Therefore, option b) results in the emission of the highest-energy photon.

To determine which transition results in the emission of the highest-energy photon in the Bohr hydrogen atom, we need to compare the energy levels associated with each transition.

The energy levels in the Bohr model of the hydrogen atom are given by the formula:
E = -13.6 eV/n^2

Where E is the energy level, n is the principal quantum number, and -13.6 eV is a constant representing the ionization energy of the hydrogen atom.

To find the energy difference between two energy levels, we subtract the initial energy level from the final energy level:
ΔE = E_final - E_initial

The energy of the emitted photon is given by the energy difference between the initial and final energy levels:
E_photon = ΔE

Let's calculate the energy differences for each transition:

a. n=1, n=6
ΔE = (E_final - E_initial) = E_6 - E_1 = (-13.6 eV/6^2) - (-13.6 eV/1^2) = -0.15 eV

b. n=6, n=1
ΔE = (E_final - E_initial) = E_1 - E_6 = (-13.6 eV/1^2) - (-13.6 eV/6^2) = -0.15 eV

c. n=6, n=3
ΔE = (E_final - E_initial) = E_3 - E_6 = (-13.6 eV/3^2) - (-13.6 eV/6^2) = -0.06 eV

d. n=3, n=6
ΔE = (E_final - E_initial) = E_6 - E_3 = (-13.6 eV/6^2) - (-13.6 eV/3^2) = -0.06 eV

e. n=1, n=4
ΔE = (E_final - E_initial) = E_4 - E_1 = (-13.6 eV/4^2) - (-13.6 eV/1^2) = -0.85 eV

From the calculations, we can see that option e. (n=1, n=4) results in the highest energy difference (ΔE = -0.85 eV). Therefore, the e. (n=1, n=4) transition results in the emission of the highest-energy photon.

The most energy comes from the largest "movement" of the electron; therefore, I would think that would be from 6 to 1.