A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency v1 and ejects

Q: Why is the ground state energy of positronium different from hydrogen?

In hydrogen, the proton is so massive that the reduced mass is essentially the electron's mass. In positronium, both particles have the same mass, making the reduced mass exactly half of the electron's mass. Since energy levels are proportional to reduced mass, positronium's energy is half of hydrogen's.

Q: Why do we add the internal energy and kinetic energy of the center of mass?

Total energy of a moving composite system is the sum of its internal energy (binding energy) and the macroscopic kinetic energy of its center of mass. Both must be accounted for in the energy conservation equation.

Q: Where did the -13.6 eV come from?

It is the standard ground state energy of a hydrogen atom, derived from the Bohr model formula $E_n = -13.6/n^2$ eV.

Q: Why is the positron's initial energy zero?

The problem explicitly states that the positron is 'at rest', meaning its macroscopic kinetic energy is zero before the collision.

Q: Does the emitted photon carry away momentum?

Yes, photons carry momentum ($p = E/c$). While we ignored it here because the photon's momentum is very small compared to massive particles, a perfectly rigorous analysis would include momentum conservation as well.

Question

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $
u_1$ and ejects the electron with a kinetic energy of 10 eV. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $
u_2$. The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV. It is given that positron has the same mass as that of electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV) is ____

Sigma Percentile · Accepted Answer

### The Tale of Two Photons

Imagine you are observing a solitary hydrogen atom, resting peacefully in its ground state. Suddenly, a photon of frequency $
u_1$ strikes it. This isn't just a gentle tap; it's a high-energy collision that completely ejects the electron from the atom, sending it flying with a kinetic energy of $10 	ext{ eV}$.

This is the classic photoelectric effect applied to an atom. But the story doesn't end there. This energetic electron is about to meet its antimatter twin—a positron.

### Stage 1: The Hydrogen Atom's Sacrifice

Let's break down the first event using the principle of energy conservation. The hydrogen atom starts in its ground state, which we know has an energy of $-13.6 	ext{ eV}$. When it absorbs the incoming photon, the total initial energy becomes the sum of the atom's ground state energy and the photon's energy.

This energy is entirely transferred to the ejected electron. Therefore, we can write our master equation for the first stage:

$$h
u_1 + (-13.6 	ext{ eV}) = KE_{e^-}$$

We are given that the kinetic energy of the ejected electron is $10 	ext{ eV}$. Substituting this into our equation, we can easily solve for the energy of the first photon:

$$h
u_1 - 13.6 = 10 \implies h
u_1 = 23.6 	ext{ eV}$$

Keep this value safe; it is the first piece of our puzzle.

### Stage 2: The Birth of Positronium

Now, our $10 	ext{ eV}$ electron is hurtling through space until it encounters a stationary positron. A positron is the exact anti-particle of an electron—it has the same mass $m$, but a positive charge. When they collide, they don't just bounce off each other; they bind together to form an exotic, hydrogen-like atom called **Positronium**.

Here is where a beautiful piece of physics comes into play. In a normal hydrogen atom, the proton is so massive that we can assume it stays perfectly still while the electron orbits it. But in Positronium, both particles have the exact same mass. They must orbit their common center of mass.

To calculate the energy levels of this new atom, we cannot use the standard electron mass. We must use the **reduced mass** ($\mu$) of the system:

$$\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$$

Since the energy levels of a Bohr atom are directly proportional to the reduced mass, the ground state energy of Positronium will be exactly half that of a Hydrogen atom:

$$E_{	ext{ps}} = \frac{-13.6 	ext{ eV}}{2} = -6.8 	ext{ eV}$$

### The Final Energy Balance

As the electron and positron bind to form this ground-state Positronium, they release excess energy by emitting a second photon of frequency $
u_2$. Additionally, the entire Positronium atom moves as a single unit with a center-of-mass kinetic energy of $5 	ext{ eV}$.

Let's apply energy conservation to this spectacular collision. The initial energy of the system is just the kinetic energy of the incoming electron (since the positron was at rest). The final energy is the sum of the Positronium's internal binding energy, its center-of-mass kinetic energy, and the energy of the emitted photon.

$$KE_{e^-} + KE_{e^+} = E_{	ext{ps}} + KE_{	ext{cm}} + h
u_2$$

Substituting our known values:

$$10 + 0 = -6.8 + 5 + h
u_2$$

Now, it's just simple arithmetic. $-6.8 + 5$ gives us $-1.8$. Moving that to the other side:

$$h
u_2 = 10 + 1.8 = 11.8 	ext{ eV}$$

### The Grand Finale

We have successfully tracked the energy through both stages of this cosmic dance. The energy of the first photon was $23.6 	ext{ eV}$, and the energy of the second photon is $11.8 	ext{ eV}$.

The question asks for the difference between these two photon energies:

$$\Delta E = h
u_1 - h
u_2 = 23.6 - 11.8 = 11.8 	ext{ eV}$$

And there we have it! A seemingly complex sequence of quantum events elegantly unraveled through the unwavering law of energy conservation.

Enter Numerical Value:

Visualized Solution

$The Two-Stage Process$

$Energy Conservation: Stage 1$

$Calculating h ν_{1}$

$Positronium Atom$

$Energy of Positronium$

$Ground State Energy of Ps$

$Energy Conservation: Stage 2$

$Calculating h ν_{2}$

$Difference in Photon Energies$

$Food for Thought$

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

The Tale of Two Photons

Stage 1

Stage 2

The Final Energy Balance

The Grand Finale

Similar Questions

The recoil speed of a hydrogen atom after it emits a photon in going from $n = 5$ state to $n = 1$ state is ........ m/s.

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 $h c = 12500 eV- \overset{˚}{A}$ )

The recoil speed of a hydrogen atom after it emits a photon in going from $n = 5$ state to $n = 1$ state will be

Enter Numerical Value:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating hν1​

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Conservation: Stage 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating hν2​

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

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