The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wav

Q: Why is the second excited state $n=3$ and not $n=2$?

The lowest possible energy level is the ground state, which is $n=1$. The state immediately above it is the *first* excited state, which is $n=2$. Following this logic, the *second* excited state is $n=3$.

Q: Why is the ground state energy negative?

Energy is defined as zero when the electron is completely free and at rest at an infinite distance from the nucleus. Because the electron is bound to the nucleus by an attractive electrostatic force, you must *add* energy to free it. Therefore, its bound state energy must be less than zero, i.e., negative.

Q: Can I use the Rydberg formula directly instead of calculating energy differences?

Yes, absolutely! Since the ionization energy is $9 \times 13.6 \text{ eV}$, and the general formula is $13.6 Z^2$, you can deduce that $Z^2 = 9$. You can then plug $Z=3$ directly into the Rydberg formula $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2})$ to get the same answer.

Q: Where does the value 1240 come from in the wavelength formula?

The formula is $\lambda = \frac{hc}{E}$. Planck's constant $h$ is $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$ and the speed of light $c$ is $3 \times 10^8 \text{ m/s}$. If you multiply these and convert Joules to electron-volts ($1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$) and meters to nanometers ($1 \text{ m} = 10^9 \text{ nm}$), the constant $hc$ evaluates to approximately $1240 \text{ eV}\cdot\text{nm}$.

Q: How do I know which specific ion this question is talking about?

By comparing the given ionization energy ($9 \text{ Rydbergs}$) to the formula for a hydrogen-like ion ($Z^2 \text{ Rydbergs}$), we find $Z^2 = 9$, so $Z = 3$. The element with atomic number 3 is Lithium. Since it must be a hydrogen-like ion (having only one electron), it must be $\text{Li}^{2+}$.

Question

The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

Options:

(A) 8.6 nm

(B) 24.2 nm

(C) 11.4 nm

(D) 35.8 nm

Sigma Percentile · Accepted Answer

### Decoding the Hydrogen-Like Ion

Imagine a mysterious hydrogen-like ion. We are told that the energy required to ionize it from its ground state is $9 	ext{ Rydbergs}$. What does this mean physically? Ionization is the process of completely removing an electron from the atom, taking it from its lowest energy state ($n=1$) all the way to infinity ($n=\infty$).

The energy required for this journey is exactly the magnitude of the ground state energy. Therefore, the ground state energy, $E_1$, is simply $-9 	ext{ Rydbergs}$. Since we know that $1 	ext{ Rydberg}$ is equivalent to $13.6 	ext{ eV}$, we can express the ground state energy as:

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

For any hydrogen-like species, the energy of the $n^{	ext{th}}$ orbit scales inversely with the square of the principal quantum number, $n$. This gives us a powerful master equation for the energy levels of this specific ion:

$$E_n = \frac{E_1}{n^2} = \frac{-9 	imes 13.6}{n^2} 	ext{ eV}$$

### Analyzing the Transition

The problem asks us to find the wavelength of the photon emitted when the electron jumps from the **second excited state** to the **ground state**.

**Don't make a silly mistake here!** The ground state is $n=1$. The *first* excited state is $n=2$. Therefore, the *second* excited state corresponds to $n=3$. The electron is making a quantum leap from $n=3$ down to $n=1$.

The energy of the emitted photon, $\Delta E$, is the difference between the initial and final energy levels:

$$\Delta E = E_3 - E_1$$

Let's substitute our energy formula into this expression:

$$\Delta E = \left( \frac{-9 	imes 13.6}{3^2} ight) - \left( \frac{-9 	imes 13.6}{1^2} ight)$$

Notice the beautiful mathematical symmetry here. In the first term, the $9$ in the numerator perfectly cancels with the $3^2$ in the denominator. Factoring out the common terms, we get:

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

$$\Delta E = 8 	imes 13.6 	ext{ eV}$$

### The Final Calculation

We now have the energy of the photon, but we need its wavelength. We can use the standard Planck-Einstein relation, $\lambda = \frac{hc}{\Delta E}$. For quick and accurate calculations in modern physics, it is highly recommended to use the approximation $hc \approx 1240 	ext{ eV}\cdot	ext{nm}$.

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

Substituting our calculated energy:

$$\lambda = \frac{1240}{8 	imes 13.6}$$

$$\lambda \approx 11.39 	ext{ nm}$$

Rounding to one decimal place, we get **$11.4 	ext{ nm}$**, which perfectly matches option (c).

### An Alternative Path: The Rydberg Formula

Could we have solved this differently? Absolutely! The ionization energy of a hydrogen-like ion is given by $13.6 Z^2 	ext{ eV}$. Since we are given that the ionization energy is $9 	imes 13.6 	ext{ eV}$, we can immediately deduce that $Z^2 = 9$, which means $Z = 3$. This mysterious ion is actually $	ext{Li}^{2+}$!

Armed with $Z=3$, you could directly apply the Rydberg wavelength formula:

$$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$$

Substituting $Z=3$, $n_f=1$, and $n_i=3$ will lead you to the exact same elegant result.

Animated Solution for Physics - Atoms and Nuclei: The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

Select Answer:

Visualized Solution

\text{Ionisation Energy}

\text{Ground State Energy}

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

\text{The Transition}

\text{Energy of Emitted Photon}

\text{Calculating } \Delta E

\text{Wavelength Formula}

\text{Final Calculation}

\text{Alternative Approach}

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

Decoding the Hydrogen-Like Ion

Analyzing the Transition

The Final Calculation

An Alternative Path

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