Q.A $12.5\ \text{eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Concept: Bohr Model Energy Levels — a $12.5\ \text{eV}$ electron beam can only excite hydrogen atoms (ground state $E_1=-13.6\ \text{eV}$) to levels whose energy gap it can fully supply.

Reasoning:

1. $\Delta E_{1\to2} = 10.2\ \text{eV}$ and $\Delta E_{1\to3} \approx 12.09\ \text{eV}$ are both $\le 12.5\ \text{eV}$ — reachable. $\Delta E_{1\to4} = 12.75\ \text{eV}$ is not — $n=4$ is out of reach.
2. So atoms get excited to $n=2$ or $n=3$ only.
3. Downward transitions: $3\to1$, $2\to1$ (both Lyman series, UV), and $3\to2$ (Balmer series, the $H_\alpha$ red line) — three lines total. …

Why the Bohr model is the right tool

At room temperature, essentially all hydrogen atoms sit in the ground state ($n=1$). When the 12.5 eV electron beam collides with these atoms, a beam electron can transfer some of its kinetic energy to the bound atomic electron — but only in the exact discrete amounts that match the gap between two Bohr energy levels. If the beam energy is short of the next gap, no excitation happens (the collision is elastic). So the first job is to find which excited levels are actually reachable.

The hydrogen energy levels are:

Step 1 — Which levels can 12.5 eV reach?

Ground state: $E_1 = -13.6\ \text{eV}$.

The 12.5 eV beam can supply $10.2\ \text{eV}$ (reaching $n=2$) and $12.09\ \text{eV}$ (reaching $n=3$), but not $12.75\ \text{eV}$ (reaching $n=4$). So the highest level any atom can be excited to is $n=3$; the beam electron that caused a $1\to3$ excitation keeps the leftover $12.5 - 12.09 = 0.41\ \text{eV}$ as its own kinetic energy — the atom only ever absorbs a whole discrete quantum, never a fraction.

Step 2 — List every allowed downward transition

Some atoms end up excited to $n=2$, some to $n=3$. Each can then fall to any lower level:

• From $n=3$: $3\to2$ and $3\to1$
• From $n=2$: $2\to1$

That gives three distinct spectral lines in total (an atom excited to $n=3$ may cascade $3\to2\to1$, emitting two photons, or jump directly $3\to1$, emitting one — across many atoms, all three lines appear).

Step 3 — Compute each wavelength

Using the Rydberg relation $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ with $R = 1.097 \times 10^7\ \text{m}^{-1}$:

$3\to1$ (Lyman):

$2\to1$ (Lyman): …

Method: Energy-Level Transition Analysis Using the Bohr Model

The core idea is that an incoming electron can transfer only discrete amounts of energy to a hydrogen atom — exactly the gaps between Bohr energy levels. Once the atom is excited, it de-excites in steps, emitting photons whose wavelengths correspond to the Lyman, Balmer, or Paschen series.

Step 1: Write the Bohr energy levels for hydrogen

For hydrogen ($Z = 1$), the energy of the $n$-th orbit is:

So the first few levels are:

Step 2: Find the maximum excitation possible

The incoming electron has $12.5\ \text{eV}$ of kinetic energy. A ground-state hydrogen atom ($E_1 = -13.6\ \text{eV}$) can absorb at most this much energy. The highest energy level reachable is the one where:

Check $n=3$: $E_3 - E_1 = (-1.51) - (-13.6) = 12.09\ \text{eV}$ — this is less than $12.5\ \text{eV}$, so $n=3$ is reachable.

Check $n=4$: $E_4 - E_1 = (-0.85) - (-13.6) = 12.75\ \text{eV}$ — this exceeds $12.5\ \text{eV}$, so $n=4$ is not reachable.

Thus the atom can be excited to $n=2$ or $n=3$ only.

Step 3: List all possible downward transitions

From $n=3$:

• $3 \to 2$ (Balmer series)
• $3 \to 1$ (Lyman series) …

Students often lose marks on this exact problem because they rush past two key ideas: the energy levels of hydrogen and the meaning of "room temperature". Let me walk through the mistakes one by one.

Mistake 1: Forgetting that hydrogen at room temperature is in the ground state

At room temperature, almost all hydrogen atoms are in $n=1$. Students sometimes assume atoms are already excited, or they try to use the Maxwell-Boltzmann distribution unnecessarily. The exam expects you to know: room temperature means $k_B T \approx 0.025\ \text{eV}$, far too small to excite hydrogen from $n=1$ to $n=2$ (which needs $10.2\ \text{eV}$). So every atom starts at $n=1$.

How to avoid: Always check the initial state. If the problem says "gaseous hydrogen at room temperature", the answer is $n=1$ for all atoms. Write it down explicitly before doing anything else.

Mistake 2: Using the beam energy as the exact excitation energy

The electron beam has $12.5\ \text{eV}$ of kinetic energy. Students often think this means the atom absorbs exactly $12.5\ \text{eV}$ and jumps to some level. But an electron can transfer any amount up to its full kinetic energy — it doesn't have to give all of it. The atom can absorb $10.2\ \text{eV}$ (to reach $n=2$) or $12.09\ \text{eV}$ (to reach $n=3$), and the leftover energy stays with the scattered electron.

How to avoid: List the excitation energies from $n=1$:

• $n=1 \to 2$: $13.6 - 3.4 = 10.2\ \text{eV}$
• $n=1 \to 3$: $13.6 - 1.51 = 12.09\ \text{eV}$
• $n=1 \to 4$: $13.6 - 0.85 = 12.75\ \text{eV}$

Now compare with $12.5\ \text{eV}$. The atom can reach $n=2$ and $n=3$ (both below $12.5\ \text{eV}$), but not $n=4$ (needs $12.75\ \text{eV} > 12.5$). So the maximum excited state is $n=3$.

Mistake 3: Only listing one transition per atom

After excitation, the atom de-excites. A single atom excited to $n=3$ can return to ground via multiple paths: $3\to2$, $2\to1$, or directly $3\to1$. Each path emits a photon of a specific wavelength. Students sometimes list only the direct $3\to1$ transition, forgetting the cascade.

How to avoid: For each excited state reached, list all possible downward transitions. For $n=3$:

• $3 \to 2$ (Balmer series, visible)
• $2 \to 1$ (Lyman series, UV)
• $3 \to 1$ (Lyman series, UV)

For $n=2$ (also reached by some atoms):

• $2 \to 1$ (Lyman series)

So the emitted wavelengths come from three distinct transitions: $3\to2$, $2\to1$, and $3\to1$. That's three different wavelengths.

Mistake 4: Confusing series names with specific lines

Students sometimes say "Lyman series will be emitted" without specifying which lines. The question asks "What series of wavelengths?" — the answer is: two lines from the Lyman series ($2\to1$ and $3\to1$) and one line from the Balmer series ($3\to2$). Don't just name the series; mention how many lines in each. …