Consider a hydrogen atom with its electron in the nth orbit. An electromagnetic radiation of wavelen

Q: Why do we use $1242 \text{ eV nm}$ for $hc$ instead of standard SI units?

Using $hc = 1242 \text{ eV nm}$ is a massive time-saver in modern physics problems! It allows you to directly plug in the wavelength in nanometers and get the energy in electron-volts, bypassing tedious conversions with Joules and meters.

Q: What exactly is ionization energy in this context?

Ionization energy is the minimum energy required to completely remove an electron from its current orbit (state $n$) to infinity (where its potential energy is zero). It is exactly equal to the positive magnitude of the electron's total energy in that orbit.

Q: Why is the energy of the $n^{\text{th}}$ orbit negative?

The negative sign indicates that the electron is bound to the nucleus. It means you have to *add* energy to the system to free the electron. An energy of zero corresponds to a completely free electron at rest at an infinite distance.

Q: Could the electron have absorbed only part of the photon's energy?

No, this is a fundamental rule of quantum mechanics! An electron bound in an atom absorbs a photon *completely* or not at all. It's an all-or-nothing transaction.

Q: What would happen if the photon's energy was less than the ionization energy?

If the photon's energy exactly matched the energy difference between two orbits, the electron would jump to a higher orbit (excitation). If it didn't match any energy gap and was less than the ionization energy, the photon would simply pass through the atom unabsorbed.

Question

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

Sigma Percentile · Accepted Answer

Imagine you are playing a game of cosmic billiards. But instead of cue balls and eight balls, you are dealing with photons of light and electrons bound to a hydrogen atom.

This problem is a classic demonstration of the photoelectric effect applied to an isolated atom. It beautifully ties together the quantum nature of light and the quantized energy levels of the Bohr model.

Let's dive into the mechanics of this quantum collision and uncover the hidden orbit of our electron!

### The Cosmic Billiards: Light Meets Matter

Our story begins with a hydrogen atom. Deep within it, an electron is orbiting the nucleus in some unknown energy level, which we call the $n^{	ext{th}}$ orbit.

Suddenly, a photon of electromagnetic radiation comes hurtling in. This photon has a specific wavelength of $90 	ext{ nm}$.

When this photon strikes the electron, it transfers all of its energy in a single, indivisible transaction. This is the hallmark of quantum mechanics—it is an all-or-nothing deal.

The energy provided by the photon is so immense that it doesn't just bump the electron to a higher orbit. It completely shatters the invisible bonds holding the electron to the nucleus.

The electron is ejected from the atom entirely, flying off into the void with a kinetic energy of $10.4 	ext{ eV}$. Our mission is to work backward from this wreckage to find out where the electron originally lived.

### The Master Equation: Conservation of Energy

To solve this mystery, we rely on one of the most unbreakable laws of the universe: the conservation of energy.

The total energy brought into the system by the incident photon must be perfectly accounted for. Where does this energy go? It splits into two distinct tasks.

First, a portion of the energy is spent paying the "exit toll." This is the Ionization Energy ($	ext{IE}_n$), the exact amount of work required to rip the electron away from the nucleus's electrostatic grip.

Second, whatever energy is left over after paying the toll is given to the electron as Kinetic Energy ($K$). This is the energy of motion that the electron uses to fly away.

We can write this elegantly as a simple equation:

$$E_{	ext{photon}} = 	ext{IE}_n + K$$

This is our master key. If we know the photon's energy and the electron's kinetic energy, we can easily find the ionization energy.

### Decoding the Incoming Photon

Before we can use our master equation, we need to know exactly how much energy the incoming photon carries.

We are given its wavelength, $\lambda = 90 	ext{ nm}$. To convert this wavelength into energy, we use the Planck-Einstein relation:

$$E_{	ext{photon}} = \frac{hc}{\lambda}$$

Here, $h$ is Planck's constant and $c$ is the speed of light.

Usually, multiplying these constants involves messy powers of ten. But the problem gives us a beautiful shortcut: $hc = 1242 	ext{ eV nm}$.

This shortcut allows us to plug in the wavelength in nanometers and instantly get the energy in electron-volts. Let's do the math:

$$E_{	ext{photon}} = \frac{1242}{90}$$

$$E_{	ext{photon}} = 13.8 	ext{ eV}$$

Our incoming photon is a concentrated packet of $13.8 	ext{ eV}$ of energy.

### Unveiling the Binding Energy

Now we have all the pieces to solve for the ionization energy. We know the photon brought in $13.8 	ext{ eV}$, and the electron flew away with $10.4 	ext{ eV}$.

Let's plug these into our conservation of energy equation:

$$13.8 = 	ext{IE}_n + 10.4$$

To find the ionization energy, we simply subtract the kinetic energy from the total photon energy:

$$	ext{IE}_n = 13.8 - 10.4$$

$$	ext{IE}_n = 3.4 	ext{ eV}$$

This tells us that it took exactly $3.4 	ext{ eV}$ of energy to break the electron free from its specific orbit. This value is the binding energy of that orbit.

### Bohr's Blueprint: Finding the Orbit

Now we must connect this binding energy to the architecture of the hydrogen atom. For this, we turn to Niels Bohr's brilliant model.

Bohr discovered that the energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom is quantized and follows a strict formula:

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

The negative sign is crucial. It means the electron is trapped in a potential well. To free it (bringing its energy to zero), you must add an amount of energy equal to the positive magnitude of $E_n$.

Therefore, the ionization energy from the $n^{	ext{th}}$ state is simply:

$$	ext{IE}_n = \frac{13.6}{n^2} 	ext{ eV}$$

We already calculated that the ionization energy for our mystery orbit is $3.4 	ext{ eV}$. Let's equate the two:

$$\frac{13.6}{n^2} = 3.4$$

Now, it is just a matter of simple algebra. We rearrange the equation to solve for $n^2$:

$$n^2 = \frac{13.6}{3.4}$$

$$n^2 = 4$$

Taking the positive square root (since orbit numbers must be positive integers), we arrive at our final destination:

$$n = 2$$

### The Grand Conclusion

The math has spoken! The electron was originally residing in the $n=2$ orbit, which is the first excited state of the hydrogen atom.

This problem is a fantastic journey from the macroscopic observation of an ejected electron down to the quantum architecture of a single atom.

By simply measuring the kinetic energy of the debris (the electron), we were able to deduce the exact initial state of the system. This very principle is the beating heart of photoelectron spectroscopy, a tool that continues to illuminate the quantum world today!

Enter Numerical Value:

Visualized Solution

$Visualizing the Process$

$E_{photon} = IE_{n} + K$

$E_{photon} = \frac{h c}{λ}$

$E_{photon} = \frac{1242}{90} eV$

$13.8 = IE_{n} + 10.4$

$IE_{n} = 13.8 - 10.4$

$E_{n} = - \frac{13.6}{n ^{2}} eV$

$\frac{13.6}{n ^{2}} = 3.4$

$n^{2} = 4 ⟹ n = 2$

\text{The Way Forward}

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

The Cosmic Billiards

The Master Equation

Decoding the Incoming Photon

Unveiling the Binding Energy

Bohr's Blueprint

The Grand Conclusion

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Process

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ephoton​=IEn​+K

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ephoton​=901242​ eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 13.8=IEn​+10.4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IEn​=13.8−10.4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​=−n213.6​ eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n213.6​=3.4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2=4⟹n=2