Step 1: Restate Bohr's three assumptions.
• The electron moves in special non-radiating circular orbits; while in such an orbit it does not lose energy even though it is accelerating.
• Only orbits with angular momentum \(L = mvr = n\hbar = \dfrac{nh}{2\pi}\) are allowed, so the orbits and their energies are quantised by the integer \(n\).
• A photon is released or absorbed only during a jump between two allowed orbits, with \(\Delta E = h\nu = \dfrac{hc}{\lambda}\).
Step 2: Level energies from the formula.
Using \(E_n = -\dfrac{13.6}{n^2}\ \text{eV}\):
\(E_2 = -3.40\ \text{eV}\), \(E_3 = -1.51\ \text{eV}\), \(E_4 = -0.85\ \text{eV}\), \(E_5 = -0.54\ \text{eV}\).
Step 3: Compute the first three Balmer photons.
All Balmer transitions land on \(n=2\).
\(3\to2:\ \Delta E = -1.51-(-3.40) = 1.89\ \text{eV}\), so \(\lambda = \dfrac{12400}{1.89} \approx 6563\ \text{Å}\) (red, Hα).
\(4\to2:\ \Delta E = -0.85-(-3.40) = 2.55\ \text{eV}\), so \(\lambda \approx 4861\ \text{Å}\) (blue-green, Hβ).
\(5\to2:\ \Delta E = -0.54-(-3.40) = 2.86\ \text{eV}\), so \(\lambda \approx 4340\ \text{Å}\) (violet, Hγ).
Step 4: How to sketch the diagram.
Stack horizontal energy levels with \(n=2\) at the bottom (\(-3.40\ \text{eV}\)) and \(n=3,4,5\) crowding closer together above it as energy rises toward \(0\). From \(n=3\), \(n=4\) and \(n=5\) draw three separate downward arrows that all terminate on the \(n=2\) line. Label them Hα, Hβ, Hγ. The arrow from the highest level (\(5\to2\)) is the longest drop and carries the most energetic (shortest wavelength) photon.
\[\boxed{\text{Three Balmer arrows: } 3\to2,\ 4\to2,\ 5\to2}\]