To determine which electron transition results in the highest-energy photon according to the Bohr model, we need to calculate the energy differences for the given transitions. The Bohr model states that the energy of an electron in a hydrogen atom is given by:
[tex]\[ E_n = -\frac{R_h c h}{n^2} \][/tex]
Where:
- [tex]\( E_n \)[/tex] is the energy at level [tex]\( n \)[/tex],
- [tex]\( R_h \)[/tex] is the Rydberg constant ([tex]\( 1.097 \times 10^7 \)[/tex] m[tex]\(^{-1}\)[/tex]),
- [tex]\( c \)[/tex] is the speed of light ([tex]\( 3 \times 10^8 \)[/tex] m/s),
- [tex]\( h \)[/tex] is Planck's constant ([tex]\( 6.626 \times 10^{-34} \)[/tex] J·s),
- [tex]\( n \)[/tex] is the principal quantum number.
The energy difference ([tex]\( \Delta E \)[/tex]) between two levels [tex]\( n_i \)[/tex] and [tex]\( n_f \)[/tex] (where [tex]\( n_i \)[/tex] is the initial level and [tex]\( n_f \)[/tex] is the final level) is given by:
[tex]\[ \Delta E = E_{n_f} - E_{n_i} = R_h c h \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \][/tex]
We need to calculate this energy difference for each given transition:
1. Transition from [tex]\( n=4 \)[/tex] to [tex]\( n=1 \)[/tex]:
[tex]\[ \Delta E_{4\to1} = R_h c h \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \][/tex]
[tex]\[ \Delta E_{4\to1} = R_h c h \left( 1 - \frac{1}{16} \right) \][/tex]
[tex]\[ \Delta E_{4\to1} = R_h c h \left( \frac{15}{16} \right) \][/tex]
2. Transition from [tex]\( n=4 \)[/tex] to [tex]\( n=3 \)[/tex]:
[tex]\[ \Delta E_{4\to3} = R_h c h \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \][/tex]
[tex]\[ \Delta E_{4\to3} = R_h c h \left( \frac{1}{9} - \frac{1}{16} \right) \][/tex]
[tex]\[ \Delta E_{4\to3} = R_h c h \left( \frac{16 - 9}{144} \right) \][/tex]
[tex]\[ \Delta E_{4\to3} = R_h c h \left( \frac{7}{144} \right) \][/tex]
3. Transition from [tex]\( n=3 \)[/tex] to [tex]\( n=4 \)[/tex]:
[tex]\[ \Delta E_{3\to4} = R_h c h \left( \frac{1}{4^2} - \frac{1}{3^2} \right) \][/tex]
[tex]\[ \Delta E_{3\to4} = R_h c h \left( \frac{1}{16} - \frac{1}{9} \right) \][/tex]
[tex]\[ \Delta E_{3\to4} = R_h c h \left( \frac{9 - 16}{144} \right) \][/tex]
[tex]\[ \Delta E_{3\to4} = R_h c h \left( \frac{-7}{144} \right) \][/tex]
Since this is a transition to a higher energy level, the energy difference is negative, indicating absorption rather than emission.
4. Transition from [tex]\( n=1 \)[/tex] to [tex]\( n=4 \)[/tex]:
[tex]\[ \Delta E_{1\to4} = R_h c h \left( \frac{1}{4^2} - \frac{1}{1^2} \right) \][/tex]
[tex]\[ \Delta E_{1\to4} = R_h c h \left( \frac{1}{16} - 1 \right) \][/tex]
[tex]\[ \Delta E_{1\to4} = R_h c h \left( \frac{1 - 16}{16} \right) \][/tex]
[tex]\[ \Delta E_{1\to4} = R_h c h \left( \frac{-15}{16} \right) \][/tex]
Again, this negative energy difference indicates absorption rather than emission.
After calculating these energy differences, we compare the magnitudes to find the highest-energy photon emission.
The energy differences for emission are:
- [tex]\( \Delta E_{4\to1} \)[/tex] (highest magnitude)
- [tex]\( \Delta E_{4\to3} \)[/tex]
The transition that results in the highest-energy photon is:
[tex]\[ n = 4 \text{ to } n = 1 \][/tex]
Thus, the transition [tex]\( n=4 \)[/tex] to [tex]\( n=1 \)[/tex] results in the highest-energy photon, with an energy of approximately [tex]\( 2.044 \times 10^{-18} \)[/tex] joules.
