Of the following transitions in the Bohr hydrogen atom, which of the following results in the emission of the lowest-energy photon.
Question 2 options:
a) n = 1 to n = 6
b) n = 5 to n = 1
c) n = 6 to n = 1
d) n = 3 to n = 5
e) n = 1 to n = 5
The energy of a photon emitted during a transition in the Bohr hydrogen atom is given by the equation:
E = -13.6 eV * (1/n_i^2 - 1/n_f^2)
Where n_i is the initial energy level and n_f is the final energy level. The higher the energy level, the lower the energy of the photon emitted.
In this case, we need to find the transition that results in the emission of the lowest-energy photon. Using the given options, we can calculate the energies of the photons emitted:
a) n = 1 to n = 6:
E = -13.6 eV * (1/1^2 - 1/6^2) = -13.6 eV * (1 - 1/36) = -13.2 eV
b) n = 5 to n = 1:
E = -13.6 eV * (1/5^2 - 1/1^2) = -13.6 eV * (1/25 - 1) = 10.88 eV
c) n = 6 to n = 1:
E = -13.6 eV * (1/6^2 - 1/1^2) = -13.6 eV * (1/36 - 1) = 12.067 eV
d) n = 3 to n = 5:
E = -13.6 eV * (1/3^2 - 1/5^2) = -13.6 eV * (1/9 - 1/25) = -2.533 eV
e) n = 1 to n = 5:
E = -13.6 eV * (1/1^2 - 1/5^2) = -13.6 eV * (1 - 1/25) = -12.944 eV
Based on these calculations, option b) (n = 5 to n = 1) results in the emission of the lowest-energy photon (-10.88 eV).
To determine which transition results in the emission of the lowest-energy photon in a Bohr hydrogen atom, we need to consider the energy levels of the different transitions.
The energy of a photon emitted during a transition in the Bohr hydrogen atom is given by the formula:
E = (Rh) * (1/nf^2 - 1/ni^2)
Where E is the energy of the photon, Rh is the Rydberg constant, and nf and ni are the final and initial energy levels, respectively.
Since we are comparing the emissions, we can ignore the negative sign and focus on the denominator of the formula. The larger the denominator, the lower the energy of the emitted photon.
Now let's evaluate the given options:
a) n = 1 to n = 6
For this transition, the denominator is 1/6^2 - 1/1^2 = 1/36 - 1 = -35/36
b) n = 5 to n = 1
The denominator in this case is 1/1^2 - 1/5^2 = 1 - 1/25 = 24/25
c) n = 6 to n = 1
The denominator for this transition is 1/1^2 - 1/6^2 = 1 - 1/36 = 35/36
d) n = 3 to n = 5
Here, the denominator is 1/5^2 - 1/3^2 = 1/25 - 1/9 = 4/225
e) n = 1 to n = 5
For this transition, the denominator is 1/5^2 - 1/1^2 = 1/25 - 1 = -24/25
Comparing the denominators, we can see that the option with the largest denominator is option (d) n = 3 to n = 5. Therefore, this transition results in the emission of the lowest-energy photon.
To determine the transition that results in the emission of the lowest-energy photon in the Bohr hydrogen atom, we need to compare the energy differences between the initial and final states.
The energy difference for the transition between two energy levels in the hydrogen atom is given by the equation:
ΔE = -13.6 eV * (1/n^2_final - 1/n^2_initial)
Let's calculate the energy differences for each option:
a) n = 1 to n = 6
ΔE = -13.6 eV * (1/6^2 - 1/1^2) = -13.6 eV * (1/36 - 1) = 13.6 eV * (35/36) = 13.33 eV
b) n = 5 to n = 1
ΔE = -13.6 eV * (1/1^2 - 1/5^2) = -13.6 eV * (1 - 1/25) = 13.6 eV * (24/25) = 13.06 eV
c) n = 6 to n = 1
ΔE = -13.6 eV * (1/1^2 - 1/6^2) = -13.6 eV * (1 - 1/36) = 13.6 eV * (35/36) = 13.33 eV
d) n = 3 to n = 5
ΔE = -13.6 eV * (1/5^2 - 1/3^2) = -13.6 eV * (1/25 - 1/9) = 13.6 eV * (8/225) = 0.48 eV
e) n = 1 to n = 5
ΔE = -13.6 eV * (1/5^2 - 1/1^2) = -13.6 eV * (1/25 - 1) = 13.6 eV * (24/25) = 13.06 eV
Comparing the energy differences calculated above, the lowest-energy photon emission occurs during the transition from n = 3 to n = 5, option d) n = 3 to n = 5.