A hydrogen-like atom (described by the Bohrs model) is observed to emit six wavelengths, originating

Q: Why do we use the formula $\frac{n(n-1)}{2}$ for the number of wavelengths?

When an electron transitions between $n$ different energy levels, it can jump between any pair of levels. The number of unique pairs (and thus unique wavelengths) is simply the number of ways to choose 2 levels out of $n$, which is given by the combination formula $^nC_2 = \frac{n(n-1)}{2}$.

Q: How do we know which energy corresponds to which level?

Energy levels in an atom are negative, and they get closer to zero as the principal quantum number $n$ increases. Therefore, the more negative value ($-0.85$ eV) corresponds to the lower energy level ($m$), and the less negative value ($-0.544$ eV) corresponds to the higher energy level ($m+3$).

Q: Why is the energy of the electron negative?

The negative sign indicates that the electron is bound to the nucleus. It means you would need to supply that much positive energy to free the electron completely (bring its energy to zero at an infinite distance).

Q: How did we simplify the square root of $0.85/13.6$ without a calculator?

You can multiply the numerator and denominator by 100 to get $85/1360$. Both are divisible by 17! $85 = 17 \times 5$ and $1360 = 17 \times 80$. This simplifies to $5/80$, which is $1/16$. The square root of $1/16$ is exactly $1/4$ or $0.25$.

Q: Why does the smallest wavelength correspond to the maximum energy difference?

According to the Planck-Einstein relation, $E = \frac{hc}{\lambda}$. Energy and wavelength are inversely proportional. Therefore, the transition that releases the most energy will produce the photon with the shortest wavelength.

Question

A hydrogen-like atom (described by the Bohrs model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between $– 0.85 	ext{ eV}$ and $–0.544 	ext{ eV}$ (including both these values).
(a) Find the atomic number of the atom.
(b) Calculate the smallest wavelength emitted in these transitions.
(Take $hc = 1240 	ext{ eV-nm}$, ground state energy of hydrogen atom $= -13.6 	ext{ eV}$)

Sigma Percentile · Accepted Answer

### Decoding the Spectral Lines

Imagine an electron in a hydrogen-like atom jumping between different energy levels, like a person hopping down the steps of a ladder. Every time it jumps down, it releases a burst of energy in the form of a photon. The problem tells us that we observe exactly **six distinct wavelengths**.

How does this help us? Well, if the electron is moving between $n$ different energy levels, the number of possible unique jumps (and thus unique wavelengths) is given by the combination formula for choosing 2 levels out of $n$:

$$ 	ext{Total lines} = \frac{n(n-1)}{2} $$

Setting this equal to 6, we get $n(n-1) = 12$, which perfectly solves to **$n = 4$**. This means our electron is playing around on exactly four consecutive steps of the energy ladder.

### The Energy Boundaries

Let's label these four consecutive energy levels. If the lowest level is $m$, the highest level in this group must be $m+3$. The problem provides the energy values for the boundaries of this group:

*   Lowest energy (most negative): $E_m = -0.85 	ext{ eV}$
*   Highest energy (least negative): $E_{m+3} = -0.544 	ext{ eV}$

### Bohr's Master Equation

To connect these energies to the atomic number $z$ and the principal quantum numbers, we bring in the heavy artillery—Bohr's energy formula for hydrogen-like atoms:

$$ E_n = -13.6 \frac{z^2}{n^2} 	ext{ eV} $$

Let's apply this formula to our two boundary levels. For the $m$-th level:

$$ -13.6 \frac{z^2}{m^2} = -0.85 \implies \frac{z^2}{m^2} = \frac{0.85}{13.6} $$

Here is a neat math trick: multiply the numerator and denominator by 100 to get $85/1360$. Both are divisible by 17, simplifying beautifully to $1/16$. Taking the square root gives us:

$$ \frac{z}{m} = \frac{1}{4} = 0.25 $$

Similarly, for the $(m+3)$-th level:

$$ -13.6 \frac{z^2}{(m+3)^2} = -0.544 \implies \frac{z^2}{(m+3)^2} = \frac{0.544}{13.6} $$

This simplifies to $1/25$. Taking the square root yields:

$$ \frac{z}{m+3} = \frac{1}{5} = 0.2 $$

### Solving the Quantum Puzzle

We now have a simple system of linear equations:
1.  $z = 0.25 m$
2.  $z = 0.2 (m + 3)$

Equating the two expressions for $z$:

$$ 0.25 m = 0.2 m + 0.6 \implies 0.05 m = 0.6 \implies \mathbf{m = 12} $$

Plugging $m = 12$ back into the first equation gives $z = 0.25 	imes 12 = \mathbf{3}$.

The atomic number is 3, which means we are dealing with a doubly ionized Lithium atom ($Li^{2+}$)!

### The Quest for the Smallest Wavelength

For the second part of the problem, we need to find the smallest wavelength emitted. According to the Planck-Einstein relation ($E = \frac{hc}{\lambda}$), wavelength is inversely proportional to energy. Therefore, the **smallest wavelength** corresponds to the **maximum energy transition**.

The largest energy gap in our group of four levels is the jump from the very top ($m+3$) to the very bottom ($m$). Let's calculate this energy difference:

$$ \Delta E_{max} = E_{m+3} - E_m = -0.544 - (-0.85) = \mathbf{0.306 	ext{ eV}} $$

Finally, we calculate the wavelength:

$$ \lambda_{min} = \frac{hc}{\Delta E_{max}} = \frac{1240 	ext{ eV-nm}}{0.306 	ext{ eV}} = \mathbf{4052.3 	ext{ nm}} $$

And there we have it! A beautiful interplay of combinatorics, algebra, and quantum mechanics.

Visualized Solution

$Number of Spectral Lines$

$Energy Boundaries$

$Bohr Energy Formula$

$Setting up Equations$

$Solving for z and m$

$Smallest Wavelength$

$Calculating λ_{min}$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Decoding the Spectral Lines

The Energy Boundaries

Bohr's Master Equation

Solving the Quantum Puzzle

The Quest for the Smallest Wavelength

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Spectral Lines

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Boundaries

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bohr Energy Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for z and m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Smallest Wavelength

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating λmin​