The K_alpha X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K

Q: Why do we subtract the photon energy from the K-hole state energy?

When an electron drops from the L-shell to the K-shell, the atom transitions from a higher energy state (K-hole) to a lower energy state (L-hole). The difference in energy is released as the photon. Therefore, $E_{\text{L-hole}} = E_{\text{K-hole}} - E_{\text{photon}}$.

Q: I'm confused about 'K-hole' and 'L-hole' states. Aren't we talking about electron energies?

It's easier to think about the total energy of the atom. Removing an inner K-electron requires more energy than removing an outer L-electron. So, an atom with a missing K-electron (K-hole state) has higher total energy than an atom with a missing L-electron (L-hole state).

Q: Why is the value of $h$ given in $\text{eV}\cdot\text{s}$ instead of $\text{J}\cdot\text{s}$?

It's a massive time-saver! Since the final answer and the given K-hole energy are in $\text{keV}$, using $h$ in $\text{eV}\cdot\text{s}$ allows you to calculate the photon energy directly in electron-volts, skipping the tedious conversion from Joules.

Q: What if the question asked for the $K_\beta$ X-ray?

A $K_\beta$ X-ray is emitted when an electron from the M-shell drops to the K-shell. You would use the energy of the M-hole state instead of the L-hole state: $E_{K_\beta} = E_K - E_M$.

Q: Is the energy of the L-hole state the same as the binding energy of an L electron?

Yes, essentially. The energy required to knock out an L electron from a neutral atom is the binding energy of that L electron, which leaves the atom in the L-hole state.

Question

The $K_{\alpha}$ X-ray of molybdenum has wavelength $0.071	ext{ nm}$. If the energy of a molybdenum atom with a K electron knocked out is $27.5	ext{ keV}$, the energy of this atom when an L electron is knocked out will be ......... keV.
(Round off to the nearest integer)
[Take, $h = 4.14 	imes 10^{-15}	ext{ eV-s}$, $c = 3 	imes 10^8	ext{ ms}^{-1}$]

Sigma Percentile · Accepted Answer

### The Mystery of the Missing Electron

Imagine a bustling city, which is our molybdenum atom. The electrons are the citizens, living in different concentric rings or "shells" around the city center (the nucleus). The innermost ring is the K-shell, the next is the L-shell, and so on.

Now, suppose a high-energy particle crashes into our city and completely knocks out one of the VIP citizens from the innermost K-shell. The atom is now in a state of high stress and energy. We call this the **K-hole state**. The problem tells us that the energy of the atom in this highly agitated state is $27.5 	ext{ keV}$.

### The Great Migration

Nature hates instability. To calm the atom down, an electron from the neighboring L-shell decides to jump down and fill the empty spot in the K-shell. But here is the catch: the L-shell electron had more energy where it was. To move to the lower-energy K-shell, it must throw away its excess energy.

It does this by emitting a brilliant flash of light—an X-ray photon! Because this photon comes from an L-to-K transition, we call it a **$K_\alpha$ X-ray**.

After this jump, the K-shell is full again, but now there is a missing electron in the L-shell. The atom has transitioned into the **L-hole state**. Our goal is to find the energy of the atom in this new state, let's call it $E_L$.

### Decoding the Photon's Message

The energy carried away by the $K_\alpha$ photon is exactly equal to the difference in the atom's energy before and after the jump. Mathematically, we can write this beautiful conservation of energy as:

$$E_{K_\alpha} = E_K - E_L$$

We don't know the photon's energy directly, but we have its fingerprint: its wavelength $\lambda = 0.071 	ext{ nm}$. We can translate this wavelength into energy using the legendary Planck-Einstein relation:

$$E_{K_\alpha} = \frac{hc}{\lambda}$$

Let's plug in the numbers. The problem kindly provides Planck's constant $h$ in electron-volts, which saves us a massive headache!

$$E_{K_\alpha} = \frac{(4.14 	imes 10^{-15} 	ext{ eV}\cdot	ext{s}) 	imes (3 	imes 10^8 	ext{ m/s})}{0.071 	imes 10^{-9} 	ext{ m}}$$

Multiplying the top gives $12.42 	imes 10^{-7}$. Dividing by the bottom gives:

$$E_{K_\alpha} \approx 17.5 	imes 10^3 	ext{ eV} = 17.5 	ext{ keV}$$

### The Final Reveal

Now we have all the pieces of the puzzle. The atom started with $27.5 	ext{ keV}$ of energy. It threw away $17.5 	ext{ keV}$ in the form of an X-ray photon.

So, what is left?

$$E_L = E_K - E_{K_\alpha}$$

$$E_L = 27.5 	ext{ keV} - 17.5 	ext{ keV} = 10 	ext{ keV}$$

The energy of the molybdenum atom when an L electron is knocked out is exactly **$10 	ext{ keV}$**. It is a beautiful demonstration of how the microscopic world perfectly balances its energy checkbook!

Enter Numerical Value:

Visualized Solution

$X -ray Emission Mechanism$

$E n er g y of K_{α} Photon$

$S u b s t i t u t in g Values$

$C a l c u l a t in g Photon Energy$

$E n er g y of L-hole State$

$F u r t h er Exploration$

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

The Mystery of the Missing Electron

The Great Migration

Decoding the Photon's Message

The Final Reveal

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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; } X-ray Emission Mechanism

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of Kα​ Photon

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Photon Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of L-hole State

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