In Li++, electron in first Bohr orbit is excited to a level by a radiation of wavelength lambda. Whe

Q: Why do we use $Z=3$ for the $\text{Li}^{++}$ ion?

The atomic number $Z$ represents the number of protons in the nucleus. Lithium always has 3 protons, regardless of how many electrons it has lost. Since $\text{Li}^{++}$ has lost two electrons, it only has one left, making it a 'hydrogen-like' ion where Bohr's model perfectly applies.

Q: Where does the formula $N = \frac{n(n-1)}{2}$ come from?

It comes from combinations. If an electron is in the $n^{\text{th}}$ state, it can transition between any two levels from $1$ to $n$. The number of unique pairs of levels is simply 'n choose 2', which is mathematically written as $^nC_2 = \frac{n(n-1)}{2}$.

Q: Why is the energy of the orbits negative?

The negative sign indicates a bound state. It means the electron is trapped by the attractive force of the nucleus. You would have to supply positive energy (equal to the magnitude of the bound energy) to completely free the electron from the atom.

Q: Can we use $hc = 1242 \text{ eV}\cdot\text{nm}$ instead of $1240$?

Yes, both are common approximations. $1242$ is slightly more accurate, but $1240$ often makes manual calculations easier because it has more factors. In multiple-choice questions, both will lead you to the same closest option.

Q: What does 'all possible ways' mean for de-excitation?

It means we are considering a large sample of $\text{Li}^{++}$ ions. Some electrons might jump directly from $n=4$ to $n=1$. Others might jump $4 \rightarrow 3$, then $3 \rightarrow 2$, then $2 \rightarrow 1$. 'All possible ways' ensures we count every unique photon wavelength emitted across the entire sample.

Question

In $	ext{Li}^{++}$, electron in first Bohr orbit is excited to a level by a radiation of wavelength $\lambda$. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $\lambda$?
[Take, $h = 6.63 	imes 10^{-34} 	ext{ Js}$; $c = 3 	imes 10^8 	ext{ ms}^{-1}$]

Options:

(A) 9.4 nm

(B) 12.3 nm

(C) 10.8 nm

(D) 11.4 nm

Sigma Percentile · Accepted Answer

Imagine you are looking at a tiny, highly energetic system: a $	ext{Li}^{++}$ ion. This ion is essentially a lithium atom that has been stripped of two of its electrons, leaving just one lonely electron orbiting a nucleus with a charge of $+3e$.

Initially, this electron is resting comfortably in its ground state, the first Bohr orbit ($n=1$). Suddenly, a photon of wavelength $\lambda$ strikes it. The electron absorbs this photon's energy and is catapulted into a higher energy level, $n$.

### Analyzing the Spectral Lines

When the electron eventually falls back down to the ground state, it doesn't always take a direct route. It can cascade down through intermediate energy levels, emitting a photon at each step. The problem tells us that a total of 6 distinct spectral lines are observed during this de-excitation process.

We have a beautiful combinatorial formula to determine the total number of possible spectral lines $N$ when an electron drops from the $n^{	ext{th}}$ state to the ground state:

$$N = \frac{n(n-1)}{2}$$

By substituting the given value of $N = 6$, we can easily find the excited state $n$:

$$6 = \frac{n(n-1)}{2}$$

$$12 = n(n-1)$$

Solving this simple quadratic equation yields **$n = 4$**. This tells us that the incident photon had exactly enough energy to excite the electron from $n=1$ to $n=4$.

### The Master Equation for Energy

Now that we know the initial and final states, we need to calculate the energy difference between them. The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like ion is given by Bohr's model:

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

For our $	ext{Li}^{++}$ ion, the atomic number is $Z = 3$. Let's calculate the energy of the ground state ($n=1$) and the excited state ($n=4$).

The energy of the ground state is:

$$E_1 = -13.6 	imes \frac{3^2}{1^2} = -13.6 	imes 9 	ext{ eV}$$

The energy of the $3^{	ext{rd}}$ excited state ($n=4$) is:

$$E_4 = -13.6 	imes \frac{3^2}{4^2} = -\frac{13.6 	imes 9}{16} 	ext{ eV}$$

### Calculating the Energy Difference

The energy absorbed by the electron, $\Delta E$, is simply the difference between these two energy levels:

$$\Delta E = E_4 - E_1$$

$$\Delta E = \left(-\frac{13.6 	imes 9}{16}ight) - (-13.6 	imes 9)$$

To make the calculation elegant, let's factor out $13.6 	imes 9$:

$$\Delta E = 13.6 	imes 9 \left(1 - \frac{1}{16}ight)$$

$$\Delta E = 13.6 	imes 9 	imes \frac{15}{16} 	ext{ eV}$$

### Final Calculation for Wavelength

This energy difference $\Delta E$ must exactly equal the energy of the incident photon. The relationship between a photon's energy and its wavelength is $\Delta E = \frac{hc}{\lambda}$.

To speed up our calculations, we use the incredibly handy approximation $hc \approx 1240 	ext{ eV}\cdot	ext{nm}$. Rearranging for $\lambda$ gives:

$$\lambda = \frac{1240}{\Delta E}$$

Substituting our expression for $\Delta E$:

$$\lambda = \frac{1240}{13.6 	imes 9 	imes \frac{15}{16}}$$

$$\lambda = \frac{1240 	imes 16}{13.6 	imes 9 	imes 15}$$

Carefully evaluating this fraction yields:

**$\lambda \approx 10.8 	ext{ nm}$**

This perfectly matches option (c). The beauty of this problem lies in how it seamlessly connects the macroscopic observation of spectral lines to the microscopic quantum jumps of a single electron.

Select Answer:

Visualized Solution

\text{Energy Levels of } \text{Li}^{++}

\text{Number of Spectral Lines}

\text{Finding the Excited State } n

\text{Energy of } n^{\text{th}} \text{ Orbit}

\text{Energy of States } n=1 \text{ and } n=4

\text{Energy Difference } \Delta E

\text{Wavelength of Incident Photon}

\text{Calculating } \lambda

\text{Extensions and Variations}

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

Analyzing the Spectral Lines

The Master Equation for Energy

Calculating the Energy Difference

Final Calculation for Wavelength

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