A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colli

Q: Why doesn't the second photon hit the atom while it's still in the excited state?

The lifetime of an electron in an excited state is extremely short, typically on the order of $10^{-8}$ seconds (10 nanoseconds). The problem states the second photon arrives after a microsecond ($10^{-6}$ seconds), which is 100 times longer! By the time the second photon arrives, the electron has already returned to the ground state.

Q: What would happen if the second photon arrived within a nanosecond?

If the second photon arrived while the electron was still in the $n=2$ state, the ionization energy required would only be 3.4 eV. The 15 eV photon would then eject the electron with a much higher kinetic energy of $15 - 3.4 = 11.6$ eV.

Q: Why does the first photon get perfectly absorbed?

A photon can only be absorbed by an atom if its energy exactly matches the energy difference between two allowed quantum states. Since $E_2 - E_1 = -3.4 - (-13.6) = 10.2$ eV, the 10.2 eV photon is a perfect match for this transition.

Q: Where does the emitted 10.2 eV photon come from?

When the electron drops from the higher energy state ($n=2$) back to the lower energy state ($n=1$), it must release the extra energy it had absorbed. This energy is released in the form of a photon with exactly 10.2 eV of energy.

Q: Why is the final kinetic energy of the electron exactly 1.4 eV?

To completely remove the electron from the ground state (ionization), 13.6 eV of energy is required. The incoming photon provides 15 eV. The excess energy, $15 - 13.6 = 1.4$ eV, is conserved and transforms into the kinetic energy of the newly freed electron.

Question

A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?

Options:

(A) 2 photons of energy 10.2 eV

(B) 2 photons of energy 1.4 eV

(C) One photon of energy 10.2 eV and an electron of energy 1.4 eV

(D) One photon of energy 10.2 eV and another photon of energy 1.4 eV

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a single, stationary hydrogen atom. Initially, its electron is resting peacefully in the **ground state** ($n=1$). In this state, the electron is bound to the nucleus with an energy of $-13.6 	ext{ eV}$.

Suddenly, the atom is bombarded by a photon carrying exactly $10.2 	ext{ eV}$ of energy. To understand what happens next, we need to look at the energy levels of the hydrogen atom. The energy of the first excited state ($n=2$) is $-3.4 	ext{ eV}$.

If we calculate the energy difference between these two states:
$$ \Delta E = E_2 - E_1 = -3.4 	ext{ eV} - (-13.6 	ext{ eV}) = 10.2 	ext{ eV} $$

### The First Collision

This is a perfect match! Because the incoming photon's energy exactly equals the energy gap, the atom absorbs the photon completely. The electron uses this energy to jump from the ground state to the first excited state ($n=2$).

### The Crucial Time Delay

Now, here is the catch where many students make a mistake. The problem states that the second photon arrives after a time interval on the order of a **microsecond** ($10^{-6} 	ext{ s}$).

Why is this important? An electron typically stays in an excited state for only about $10^{-8} 	ext{ seconds}$ (10 nanoseconds) before it naturally drops back down. Since a microsecond is 100 times longer than 10 nanoseconds, the electron has plenty of time to de-excite back to the ground state long before the second photon arrives!

As the electron drops back to $n=1$, it must release the energy it had previously absorbed. It does this by emitting a photon of exactly **$10.2 	ext{ eV}$**. This emitted photon flies off and will be the first thing our detector picks up.

### The Second Collision and Ionization

By the time the second photon arrives, the atom is back in its ground state. This new photon packs a much bigger punch: $15 	ext{ eV}$.

To completely free the electron from the ground state—a process called **ionization**—we need to overcome its binding energy of $13.6 	ext{ eV}$. Since $15 	ext{ eV}$ is greater than $13.6 	ext{ eV}$, the photon has more than enough energy to knock the electron completely out of the atom.

What happens to the leftover energy? According to the conservation of energy, the excess energy becomes the kinetic energy (KE) of the newly freed electron:
$$ 	ext{KE} = E_{	ext{photon}} - 	ext{Ionization Energy} $$
$$ 	ext{KE} = 15 	ext{ eV} - 13.6 	ext{ eV} = 1.4 	ext{ eV} $$

### The Final Observation

So, what does our detector ultimately see? 
1. It detects the **$10.2 	ext{ eV}$ photon** that was emitted when the atom de-excited after the first collision.
2. It detects the **ejected electron**, which is flying away with a kinetic energy of **$1.4 	ext{ eV}$**.

This perfectly matches option (c). The beauty of this problem lies in understanding the timeline of events—realizing that the atom doesn't just wait around in an excited state, but actively responds to the passage of time!

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Visualized Solution

$Initial State$

$First Collision$

$Excitation$

$Time Delay & De-excitation$

$First Observation$

$Second Collision$

$Ionization$

$Kinetic Energy of Electron$

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

Analyzing the Setup

The First Collision

The Crucial Time Delay

The Second Collision and Ionization

The Final Observation

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Delay & De-excitation

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

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

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

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