Question
The energy level diagram of a hydrogen atom shows transitions from n=4 to n=2 and from n=3 to n=2. Which transition emits a photon of larger wavelength? Find both wavelengths. Use RH=1.097×107 m−1.
Solution — Step by Step
λ1=RH(nf21−ni21)
For ni=3,nf=2 (Hα line):
λ321=1.097×107(41−91)=1.097×107×365≈1.524×106 m−1
λ32≈656 nm
λ421=1.097×107(41−161)=1.097×107×163≈2.057×106 m−1
λ42≈486 nm
λ32=656 nm (red) is larger than λ42=486 nm (blue-green).
Final answer: 3→2 transition has larger wavelength (656 nm); 4→2 gives 486 nm.
Why This Works
A larger energy gap means a higher-frequency, shorter-wavelength photon (E=hc/λ). The 4→2 jump crosses a bigger energy gap than 3→2, so its photon has shorter wavelength. Both lines lie in the Balmer series (visible light) because they end at n=2.
The Lyman series ends at n=1 (UV), Balmer at n=2 (visible), Paschen at n=3 (IR). Memorise the series names and ranges — NEET asks this every year.
Alternative Method
Compute energies and use λ=hc/ΔE:
En=−n213.6 eV
E2=−3.4 eV, E3=−1.51 eV, E4=−0.85 eV.
ΔE3→2=1.89 eV, giving λ=1240/1.89≈656 nm.
ΔE4→2=2.55 eV, giving λ=1240/2.55≈486 nm.
Use λ(nm)=1240/E(eV) as a fast shortcut.
The photon picture: the atom drops to a lower energy state and releases the difference as a photon. Energy is positive (released), but the energy levels themselves are negative because the electron is bound. Don’t confuse the sign of En with the sign of the emitted photon energy.
Common Mistake
Mixing up emission and absorption. Emission means the atom goes from higher to lower (drops down), releasing a photon. Absorption is the reverse — atom climbs up, taking energy from a photon. The wavelength formula is the same magnitude but the direction of the arrow on the diagram differs.