An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making

Q: Why do we use $Z=2$ here?

Because it's a Helium ion ($\text{He}^+$), which has an atomic number of 2. The formula requires $Z^2$.

Q: Why did we solve the second transition first?

Because we know the final state is the ground state ($n=1$), leaving only one unknown ($m$). If we started with the first transition, we would have two unknowns ($n$ and $m$).

Q: Can we add the wavelengths to get the total energy?

No, energies add up, not wavelengths. $E_{total} = E_1 + E_2$, which means $\frac{1}{\lambda_{total}} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$.

Q: What does the $13.6$ represent?

It's the ionization energy of hydrogen in eV, which serves as the base constant for Bohr's energy levels.

Q: Is the formula $\frac{1240}{\lambda}$ exact?

It's an approximation of $hc$, usually $1242$ or $1240$ depending on the given constants. The question explicitly asks to use $1240$.

Question

An excited $	ext{He}^+$ ion emits two photons in succession, with wavelengths $108.5	ext{ nm}$ and $30.4	ext{ nm}$, in making a transition to ground state. The quantum number $n$ corresponding to its initial excited state is [for photon of wavelength $\lambda$, energy $E = \frac{1240	ext{ eV}}{\lambda	ext{ (in nm)}}$]

Options:

(A) $n = 4$

(B) $n = 5$

(C) $n = 7$

(D) $n = 6$

Sigma Percentile · Accepted Answer

### The Journey of the Helium Ion

Imagine a Helium ion (`$	ext{He}^+$`) sitting high up in an excited state `$n$`. It wants to return to its ground state (`$n=1$`), but instead of making a direct jump, it decides to take a pit stop at an intermediate state `$m$`.

During this two-step journey, it emits two photons. The first photon, emitted during the jump from `$n$` to `$m$`, has a wavelength of `$\lambda_1 = 108.5	ext{ nm}$`. The second photon, emitted during the jump from `$m$` to the ground state `$1$`, has a wavelength of `$\lambda_2 = 30.4	ext{ nm}$`. Our mission is to find the initial starting point, `$n$`.

### The Master Equation

According to Bohr's model, the energy difference between any two states is given by the Rydberg formula:

$$ \Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV} $$

We are also given a handy shortcut to find the energy of a photon directly from its wavelength:

$$ E = \frac{1240}{\lambda 	ext{ (in nm)}} 	ext{ eV} $$

### Reverse Engineering: The Second Transition

There is a catch here. If we start with the first transition (`$n 	o m$`), we have two unknowns. But if we look at the second transition (`$m 	o 1$`), we only have one unknown (`$m$`), because we know the final state is the ground state (`$n_f = 1$`).

For a Helium ion, the atomic number `$Z$` is `$2$`. Let's plug the values into our master equation for the second transition:

$$ 13.6 	imes 2^2 	imes \left( \frac{1}{1^2} - \frac{1}{m^2} ight) = \frac{1240}{30.4} $$

$$ 13.6 	imes 4 	imes \left( 1 - \frac{1}{m^2} ight) = \frac{1240}{30.4} $$

Now, let's carefully calculate the right side and isolate the `$m$` term:

$$ 1 - \frac{1}{m^2} = \frac{1240}{30.4 	imes 13.6 	imes 4} \approx 0.75 $$

$$ \frac{1}{m^2} = 1 - 0.75 = 0.25 $$

$$ m^2 = \frac{1}{0.25} = 4 \implies m = 2 $$

So, the intermediate pit stop was at the first excited state, `$m=2$`.

### Finding the Origin: The First Transition

Now that we know the intermediate state is `$m=2$`, we can easily solve for the initial state `$n$` using the first transition (`$n 	o 2$`).

$$ 13.6 	imes 4 	imes \left( \frac{1}{2^2} - \frac{1}{n^2} ight) = \frac{1240}{108.5} $$

$$ \frac{1}{4} - \frac{1}{n^2} = \frac{1240}{108.5 	imes 13.6 	imes 4} $$

Let's crunch the numbers:

$$ \frac{1}{4} - \frac{1}{n^2} \approx 0.21 $$

Since `$\frac{1}{4} = 0.25$`:

$$ \frac{1}{n^2} = 0.25 - 0.21 = 0.04 $$

### The Final Calculation

We are at the finish line. If `$\frac{1}{n^2} = 0.04$`, then:

$$ n^2 = \frac{1}{0.04} = 25 $$

$$ n = 5 $$

**The initial excited state of the Helium ion was `$n=5$`.** By working backward from the known ground state, we elegantly unraveled the entire sequence of transitions!

Select Answer:

Visualized Solution

Visual Anchor

Energy Difference Formula

Second Transition ( $m \to 1$ )

Solving for $m$

First Transition ( $n \to m$ )

Solving for $n$

Final Answer

The Way Forward

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

The Journey of the Helium Ion

The Master Equation

Reverse Engineering

Finding the Origin

The Final Calculation

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