A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited

Q: Why do we add the energies of the two photons?

When an atom makes successive transitions (e.g., $n \to n_i \to 2$), the total energy lost by the atom is the sum of the energies of the photons emitted at each step. By conservation of energy, this total energy equals the energy difference between the initial and final states.

Q: What does "hydrogen-like atom" mean?

A hydrogen-like atom is any atom or ion that has only one electron, such as $\text{He}^+$, $\text{Li}^{2+}$, etc. Bohr's model applies perfectly to these systems, with the energy levels scaled by $Z^2$.

Q: Why is the first excited state $n=2$ and not $n=1$?

The ground state, which is the lowest possible energy state, is $n=1$. Any state above the ground state is an 'excited' state. Therefore, the first state above the ground state is $n=2$.

Q: Why did we subtract the two equations?

We had two equations with two unknowns ($E_n$ and $Z$). By subtracting the equation for $E_n - E_3$ from $E_n - E_2$, we eliminated the unknown $E_n$ and obtained an equation solely in terms of $Z$, which we could easily solve.

Q: Can $n$ or $Z$ be a fraction?

No, both the principal quantum number $n$ and the atomic number $Z$ must be integers. If your calculations result in a fraction, it indicates a calculation error or an invalid physical scenario.

Question

A hydrogen like atom (atomic number $Z$) is in a higher excited state of quantum number $n$. The excited atom can make a transition to the first excited state by successively emitting two photons of energy $10.2	ext{ eV}$ and $17.0	ext{ eV}$ respectively. Alternately, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies $4.25	ext{ eV}$ and $5.95	ext{ eV}$ respectively.
Determine the values of $n$ and $Z$.
(Ionization energy of H-atom $= 13.6	ext{ eV}$)

Sigma Percentile · Accepted Answer

The problem of atomic transitions often feels like a puzzle where energy levels are the missing pieces. In this thrilling journey, we will decode the secrets of a hydrogen-like atom by analyzing the photons it emits during its descent to lower energy states. Let's dive into the quantum world!

### Analyzing the Setup

We are given an atom in a higher excited state with a principal quantum number $n$. It can take two different paths to de-excite.

First, it can transition to the **first excited state**. Remember, the ground state is $n=1$, so the first excited state is $n=2$. During this transition, it emits two photons successively with energies $10.2	ext{ eV}$ and $17.0	ext{ eV}$. By the law of conservation of energy, the total energy difference between the $n$ state and the $n=2$ state is simply the sum of these photon energies.

$$E_n - E_2 = 10.2 + 17.0 = 27.2	ext{ eV}$$

Alternatively, the atom can transition from the same $n$ state to the **second excited state**, which corresponds to $n=3$. In this path, it emits two photons with energies $4.25	ext{ eV}$ and $5.95	ext{ eV}$. Again, summing these gives us the total energy difference.

$$E_n - E_3 = 4.25 + 5.95 = 10.2	ext{ eV}$$

### The Master Equation

Now we have a system of two equations. The beauty of this setup is that we can eliminate the unknown initial state energy $E_n$ by simply subtracting the second equation from the first.

$$(E_n - E_2) - (E_n - E_3) = 27.2 - 10.2$$

This simplifies elegantly to give us the energy difference between the third and second energy levels.

$$E_3 - E_2 = 17.0	ext{ eV}$$

### Finding the Atomic Number

To proceed, we need our trusty tool: **Bohr's energy formula**. For a hydrogen-like atom, the energy of the $n$-th state is given by $E_n = -13.6 \frac{Z^2}{n^2}	ext{ eV}$.

Using this formula, we can express the energy difference $E_3 - E_2$ in terms of the atomic number $Z$.

$$E_3 - E_2 = 13.6 Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$$

Let's substitute the value we found earlier and solve for $Z$.

$$13.6 Z^2 \left( \frac{1}{4} - \frac{1}{9} ight) = 17.0$$

$$13.6 Z^2 \left( \frac{5}{36} ight) = 17.0$$

Rearranging the terms to isolate $Z^2$:

$$Z^2 = \frac{17.0 	imes 36}{13.6 	imes 5} = \frac{612}{68} = 9$$

Taking the square root, we find that $Z = 3$. This reveals the identity of our mysterious atom: it is a doubly ionized Lithium atom ($	ext{Li}^{2+}$)!

### Final Calculation

With $Z$ in our toolkit, finding the initial state $n$ is a breeze. We substitute $Z=3$ back into our very first equation.

$$E_n - E_2 = 27.2	ext{ eV}$$

$$13.6 (3)^2 \left( \frac{1}{2^2} - \frac{1}{n^2} ight) = 27.2$$

Let's crunch the numbers. $13.6 	imes 9 = 122.4$.

$$122.4 \left( \frac{1}{4} - \frac{1}{n^2} ight) = 27.2$$

Dividing both sides by $122.4$:

$$\frac{1}{4} - \frac{1}{n^2} = \frac{27.2}{122.4} = \frac{2}{9}$$

Now, isolate the term with $n$:

$$\frac{1}{n^2} = \frac{1}{4} - \frac{2}{9} = \frac{9 - 8}{36} = \frac{1}{36}$$

Inverting both sides gives $n^2 = 36$, which means $n = 6$.

We have successfully unraveled the mystery! The atom is Lithium ($Z=3$), and it started its journey from the $n=6$ excited state.

Visualized Solution

$E_{n} - E_{2} = 27.2 eV$

$E_{n} - E_{3} = 10.2 eV$

$E_{3} - E_{2} = 17.0 eV$

$Δ E = 13.6 Z^{2} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$

$Z = 3$

$E_{n} - E_{2} = 27.2 eV$

$n = 6$

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

Analyzing the Setup

The Master Equation

Finding the Atomic Number

Final Calculation

Similar Questions

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Consider a hydrogen atom with its electron in the $n^{th}$ orbit. An electromagnetic radiation of wavelength $90 nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 eV$ , then the value of $n$ is ( $h c = 1242 eV nm$ )

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​−E2​=27.2 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​−E3​=10.2 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E3​−E2​=17.0 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=13.6Z2(nf2​1​−ni2​1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z=3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​−E2​=27.2 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=6

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

Analyzing the Setup

The Master Equation

Finding the Atomic Number

Final Calculation

Similar Questions

$E_{n} - E_{2} = 27.2 eV$

$E_{n} - E_{3} = 10.2 eV$

$E_{3} - E_{2} = 17.0 eV$

$Δ E = 13.6 Z^{2} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$

$Z = 3$

$E_{n} - E_{2} = 27.2 eV$

$n = 6$