In the Bohr model of the hydrogen atom, how does the energy difference between the n=4 and n=3 orbitals compare to the energy difference between the n=3 and n=2 orbitals?

In the Bohr model of the hydrogen atom, the energy difference between orbits decreases as the principal quantum number increases. This means that the energy difference between the n=4 and n=3 orbitals is actually smaller than the energy difference between the n=3 and n=2 orbitals.

To understand why, we can look at the formula for the energy levels of the hydrogen atom, which is given by:

E = -13.6 eV / n²

Here, n is the principal quantum number, and 13.6 eV is the energy of the ground state (n=1). The energy levels become less negative (and thus, higher) as n increases, which indicates that electrons in higher orbitals have more energy.

Now, we can calculate the energy for each orbital:

• For n=2: E₂ = -13.6 eV / (2²) = -3.4 eV
• For n=3: E₃ = -13.6 eV / (3²) = -1.51 eV
• For n=4: E₄ = -13.6 eV / (4²) = -0.85 eV

Now we find the energy differences:

• Energy difference between n=3 and n=2: ΔE_3,2 = E₂ – E₃ = -3.4 eV – (-1.51 eV) = -1.89 eV
• Energy difference between n=4 and n=3: ΔE_4,3 = E₃ – E₄ = -1.51 eV – (-0.85 eV) = -0.66 eV

As we can see from the calculations, the energy difference between the n=3 and n=2 orbitals is larger than that between the n=4 and n=3 orbitals. This shows how, in the Bohr model, energy levels get closer together as the value of n increases, leading to smaller energy transitions for higher energy levels.