A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 Å.

Q: Why is the ground state energy of hydrogen negative?

The negative sign indicates that the electron is bound to the nucleus. You would need to supply $13.6 \text{ eV}$ of energy (making the total energy zero) to completely free the electron from the atom's pull.

Q: Why do we add the photon's energy to the ground state energy?

This is the principle of conservation of energy. The electron absorbs the photon, meaning the photon's energy is transferred to the electron, increasing its total energy and allowing it to jump to a higher, less tightly bound orbit.

Q: What exactly is $a_0$ in the options?

$a_0$ represents the Bohr radius, which is the radius of the very first orbit ($n=1$) of a hydrogen atom. Its value is approximately $0.529 \text{ \AA}$. The radii of all other orbits are expressed as multiples of this fundamental constant.

Q: I got $n^2 = 16.19$. Why did we round it to 16?

In quantum mechanics, the principal quantum number $n$ must be an exact integer. The slight discrepancy arises because the value $13.6 \text{ eV}$ is an approximation (it's closer to $13.598 \text{ eV}$), and the given $hc$ value is also rounded. We always round to the nearest perfect square for $n^2$.

Q: Does the formula $r_n = n^2 a_0$ apply to all elements?

No, this specific formula is only for the hydrogen atom. For hydrogen-like ions (species with only one electron, like $\text{He}^+$ or $\text{Li}^{2+}$), the formula becomes $r_n = \frac{n^2}{Z} a_0$, where $Z$ is the atomic number.

Question

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 	ext{ \AA}$. The radius of the atom in the excited state in terms of Bohr radius $a_0$ will be
(Take $hc = 12500 	ext{ eV-\AA}$)

Options:

(A) $4a_0$

(B) $9a_0$

(C) $16a_0$

(D) $25a_0$

Sigma Percentile · Accepted Answer

### The Setup: A Sleeping Electron

Imagine a hydrogen atom resting peacefully in its ground state. In this state, the lone electron is in the innermost orbit, where the principal quantum number is $n = 1$. The energy of the electron in this tightly bound state is given by the well-known Bohr formula:

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

Suddenly, the tranquility is broken. A photon with a specific wavelength of $\lambda = 980 	ext{ \AA}$ strikes the atom. The electron absorbs this photon, taking in all its energy. This influx of energy will cause the electron to undergo a quantum leap to a higher, excited energy state, which we will call $n$.

### The Energy Injection

To figure out exactly where the electron goes, we first need to know how much energy the photon delivered. The energy $\Delta E$ of a photon is inversely proportional to its wavelength, described by the equation:

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

The problem kindly provides us with the value of $hc = 12500 	ext{ eV-\AA}$, which makes our calculation beautifully straightforward. Let's substitute the values:

$$\Delta E = \frac{12500}{980} \approx 12.76 	ext{ eV}$$

This $12.76 	ext{ eV}$ is the exact amount of energy the electron gains.

### Calculating the Quantum Leap

Now, we apply the principle of conservation of energy. The final energy of the electron in its excited state ($E_n$) must be the sum of its initial ground state energy and the energy it absorbed from the photon:

$$E_n = E_1 + \Delta E$$

$$E_n = -13.6 	ext{ eV} + 12.76 	ext{ eV} = -0.84 	ext{ eV}$$

We now know the energy of the new orbit. But which orbit is it? We return to the general Bohr energy formula, $E_n = \frac{-13.6}{n^2}$, and set it equal to our newly found energy:

$$\frac{-13.6}{n^2} = -0.84$$

Rearranging this to solve for $n^2$, we get:

$$n^2 = \frac{13.6}{0.84} \approx 16$$

Since $n$ must be a positive integer, we take the square root of $16$ to find that $n = 4$. The electron has leaped all the way to the fourth orbit!

### The Final Orbit Radius

The question ultimately asks for the radius of this new excited state. According to Bohr's model, the radius of the $n^{	ext{th}}$ orbit of a hydrogen atom scales with the square of the principal quantum number:

$$r_n = n^2 a_0$$

Here, $a_0$ is the Bohr radius (the radius of the ground state orbit). Since our electron is now residing in the $n = 4$ orbit, we simply substitute this value into our radius formula:

$$r_4 = (4)^2 a_0 = 16 a_0$$

**The radius of the atom in the excited state is $16 a_0$.**

Animated Solution for Physics - Atoms and Nuclei: A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 \overset{˚}{A}$ . The radius of the atom in the excited state in terms of Bohr radius $a_{0}$ will be (Take $h c = 12500 eV- \overset{˚}{A}$ )

Select Answer:

Visualized Solution

Visualizing the Setup

Energy of the Photon

Substituting Values

Calculating Photon Energy

Ground State Energy

Energy of the Excited State

Final Energy Calculation

Finding the Principal Quantum Number

Solving for n

Radius Formula

Final Answer

The Way Forward

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

The Setup

The Energy Injection

Calculating the Quantum Leap

The Final Orbit Radius

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