As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state o

Q: Why do we use the formula for hydrogen when the question is about Lithium?

Bohr's model works for any system with a single electron. A neutral Lithium atom has 3 electrons, but a doubly ionized Lithium atom ($\text{Li}^{2+}$) has lost 2 electrons, leaving exactly 1. This makes it a 'hydrogen-like' species, so Bohr's formula applies perfectly, provided we use the correct atomic number ($Z=3$).

Q: What does the negative sign in the energy formula mean?

The negative sign indicates a bound state. It means the electron is trapped in the electrostatic potential well of the nucleus. You have to supply positive energy (do work on the system) to free the electron and bring its total energy to zero.

Q: What if the electron was in the first excited state instead of the ground state?

The 'first excited state' corresponds to $n=2$. You would plug $n=2$ into the formula, giving $E_2 = -13.6 \times \frac{3^2}{2^2} = -13.6 \times \frac{9}{4} = -30.6 \text{ eV}$. The energy required to remove it would then be $30.6 \text{ eV}$.

Q: Why is $Z=3$ for $\text{Li}^{2+}$? Didn't it lose electrons?

$Z$ represents the atomic number, which is the number of protons in the nucleus. The number of protons defines the element and never changes during ionization. Lithium always has 3 protons, so $Z$ is always 3. The '2+' charge just means it has 2 fewer electrons than protons.

Q: What is the energy of the electron at infinity?

By convention, the potential energy of an electron at an infinite distance from the nucleus is taken as zero. If we supply just the minimum energy required to remove it, it will arrive at infinity with zero kinetic energy as well. Thus, its total energy at infinity is $0 \text{ eV}$.

Question

As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom ($Z = 3$) is

Options:

(A) 1.51

(B) 13.6

(C) 40.8

(D) 122.4

Sigma Percentile · Accepted Answer

This problem is a classic application of Bohr's atomic model, specifically tailored for hydrogen-like species. Let's break down the physics and the math behind it to see how elegantly simple it really is.

### Analyzing the Setup

The question asks for the minimum energy required to remove an electron from the ground state of a doubly ionized Lithium atom, denoted as $	ext{Li}^{2+}$.

First, what does "doubly ionized" mean? A neutral Lithium atom has an atomic number $Z = 3$, meaning it has 3 protons in its nucleus and 3 electrons orbiting it. When it is doubly ionized, it loses two of those electrons, leaving it with just a single electron. Any atom or ion with only one electron is called a **hydrogen-like species**. This is crucial because Bohr's model is strictly valid only for single-electron systems.

### The Master Equation

According to Bohr's model, the total energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom is given by the formula:

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

Here, $Z$ is the atomic number (number of protons) and $n$ is the principal quantum number (the orbit number). The negative sign is very important—it signifies that the electron is bound to the nucleus. It is in a "potential well," and you must supply energy to pull it out.

### Final Calculation

For our $	ext{Li}^{2+}$ ion, we know:
- The atomic number $Z = 3$.
- The electron is in the ground state, which means it is in the lowest possible energy level, so $n = 1$.

Let's substitute these values into our master equation:

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

$$E_1 = -13.6 	imes 9 	ext{ eV}$$

$$E_1 = -122.4 	ext{ eV}$$

This $-122.4 	ext{ eV}$ is the energy of the electron while it is happily orbiting the Lithium nucleus in the ground state.

The question asks for the minimum energy required to *remove* the electron. Removing the electron means taking it from its current state ($n=1$) to infinity ($n=\infty$), where it is completely free from the nucleus's pull. By convention, the energy at infinity is zero ($E_\infty = 0$).

The energy required to do this is called the **Ionization Energy**, and it is simply the difference between the final state energy and the initial state energy:

$$E_{	ext{ion}} = E_\infty - E_1$$

$$E_{	ext{ion}} = 0 - (-122.4 	ext{ eV})$$

$$E_{	ext{ion}} = +122.4 	ext{ eV}$$

So, you need to supply exactly $122.4 	ext{ eV}$ of energy to kick that electron out of the atom. This matches option (d) perfectly.

Animated Solution for Physics - Atoms and Nuclei: As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom ( $Z = 3$ ) is

Select Answer:

Visualized Solution

\text{Visualizing the } \text{Li}^{2+} \text{ Atom}

\text{Bohr's Energy Formula}

\text{Substituting the Values}

\text{Calculating Ground State Energy}

\text{Ionization Energy}

\text{Conclusion}

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

Analyzing the Setup

The Master Equation

Final Calculation

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