A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest.

Q: Why is the collision inelastic?

The collision is inelastic because some of the initial kinetic energy of the neutron is absorbed by the Helium atom to excite its electron to a higher energy state. Kinetic energy is not conserved, but total energy is.

Q: Why does the neutron's x-momentum become zero after collision?

The problem states that the neutron scatters at an angle of 90 degrees with respect to its original direction (which we took as the x-axis). Therefore, its velocity vector is entirely along the y-axis, making its x-component zero.

Q: How do we know the atom can only be excited up to n=3?

We determine this by trial and error using the energy conservation equation. When we assume excitation to n=4, the calculated kinetic energy for the neutron comes out negative, which is physically impossible. Hence, n=3 is the maximum possible excited state.

Q: Why do we use $Z=2$ for the energy level formula?

The target is a singly ionized Helium atom (He+). It has a nucleus with 2 protons, so its atomic number $Z$ is 2. Since it's singly ionized, it has only one electron left, making it a hydrogen-like species where the Bohr model formula $E_n = -13.6 Z^2/n^2$ is perfectly applicable.

Q: Why are there three possible emission frequencies?

When the electron is excited to the $n=3$ state, it can return to the ground state ($n=1$) in different ways. It can make a direct jump from $n=3$ to $n=1$, or it can jump from $n=3$ to $n=2$ and then from $n=2$ to $n=1$. Each of these three distinct jumps corresponds to a photon of a specific frequency.

Question

A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90° with respect of its original direction.
(a) Find the allowed values of the energy of the neutron and that of the atom after the collision.
(b) If the atom gets de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation.
[Given : Mass of He atom = 4 × ( mass of neutrons )
Ionization energy of H atom = 13.6 eV]

Sigma Percentile · Accepted Answer

Welcome, future physicists! Today, we are going to dive into a fascinating problem that beautifully marries classical mechanics with quantum physics. We will explore what happens when a fast-moving neutron collides with a stationary, singly ionized helium atom.

### Setting the Stage: The Collision Dynamics

Imagine a neutron, a tiny but massive particle with mass $m$, zooming in with a kinetic energy of $K = 65	ext{ eV}$. Right in its path sits a singly ionized helium atom (He$^+$) at rest. Since a helium nucleus has 2 protons and 2 neutrons, its mass is approximately $4m$.

When they collide, it's an inelastic collision. The neutron scatters at an angle of $90^\circ$ from its original path, carrying away a new kinetic energy $K_1$. The helium atom, absorbing the impact, scatters at some angle $	heta$ with kinetic energy $K_2$.

### The Math of Momentum

Since there are no external forces acting on our system, we can confidently apply the conservation of linear momentum. Let's break this down into $x$ and $y$ components.

In the $x$-direction, the initial momentum belonged entirely to the neutron. In the final state, the neutron's $x$-component is zero because it scattered at $90^\circ$. Therefore, all the $x$-momentum must be carried by the helium atom:

$$p_i = p_f \Rightarrow \sqrt{2Km} = \sqrt{2(4m)K_2} \cos	heta$$

In the $y$-direction, the initial momentum was zero. So, the final $y$-components of the neutron and the helium atom must perfectly cancel each other out:

$$\sqrt{2K_1m} = \sqrt{2(4m)K_2} \sin	heta$$

If we square and add these two equations, the $\sin^2	heta$ and $\cos^2	heta$ terms combine to give 1. This leaves us with a very neat relationship between the kinetic energies:

$$K + K_1 = 4K_2 \Rightarrow 4K_2 - K_1 = 65	ext{ eV}$$

### The Energy Equation: Where Does the Energy Go?

Now, let's talk about energy. Because the collision is inelastic, kinetic energy is not conserved. Some of the initial $65	ext{ eV}$ is spent exciting the electron in the helium atom to a higher energy state.

From the conservation of total energy, we can write:

$$K = K_1 + K_2 + \Delta E$$

$$K_1 + K_2 = 65 - \Delta E$$

### Exploring the Quantum Realm: Energy Levels of Helium

To find out how much energy the helium ion can actually absorb ($\Delta E$), we need to know its energy levels. According to Bohr's model, the energy of the $n$-th orbit for a hydrogen-like ion is:

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

For helium, $Z = 2$, so the formula becomes $E_n = -\frac{54.4}{n^2}	ext{ eV}$. Let's calculate the energies of the first few states:

$E_1 = -54.4	ext{ eV}$
$E_2 = -13.6	ext{ eV}$
$E_3 = -6.04	ext{ eV}$
$E_4 = -3.4	ext{ eV}$

The possible excitation energies from the ground state ($n=1$) are:

$\Delta E_1 = E_2 - E_1 = 40.8	ext{ eV}$
$\Delta E_2 = E_3 - E_1 = 48.36	ext{ eV}$
$\Delta E_3 = E_4 - E_1 = 51	ext{ eV}$

### The Moment of Truth: Calculating the Final Energies

Let's test these possible excitation energies.

**Case 1: Excitation to $n=2$**
If $\Delta E = 40.8	ext{ eV}$, then $K_1 + K_2 = 65 - 40.8 = 24.2	ext{ eV}$. 
Solving this simultaneously with $4K_2 - K_1 = 65	ext{ eV}$, we get:
$K_1 = 6.36	ext{ eV}$ and $K_2 = 17.84	ext{ eV}$. 
This is a perfectly valid physical state!

**Case 2: Excitation to $n=3$**
If $\Delta E = 48.36	ext{ eV}$, then $K_1 + K_2 = 65 - 48.36 = 16.64	ext{ eV}$. 
Solving the equations again yields:
$K_1 = 0.312	ext{ eV}$ and $K_2 = 16.328	ext{ eV}$. 
This is also a valid state.

### The Quantum Limit: Why Not n=4?

You might be wondering, why stop at $n=3$? Let's see what happens if we try to excite the atom to $n=4$.

If $\Delta E = 51	ext{ eV}$, then $K_1 + K_2 = 65 - 51 = 14	ext{ eV}$. 
Solving our system of equations gives $K_1 = -1.8	ext{ eV}$.

Kinetic energy can never be negative! This mathematical impossibility tells us a profound physical truth: the neutron simply does not have enough energy to excite the helium atom to the $n=4$ state while satisfying the strict laws of momentum conservation. The atom can only be excited up to $n=3$.

### The Aftermath: De-excitation and Emitted Frequencies

What goes up must come down. The excited helium atom will eventually return to its ground state, emitting photons in the process.

From the $n=3$ state, there are three possible transitions:
1. A direct jump from $n=3$ to $n=1$.
2. A jump from $n=3$ to $n=2$.
3. A subsequent jump from $n=2$ to $n=1$.

We can calculate the frequencies of these emitted photons using the relation $
u = \frac{\Delta E}{h}$:

For $n=3 ightarrow n=2$:
$$
u_1 = \frac{E_3 - E_2}{h} = \frac{-6.04 - (-13.6)}{6.63 	imes 10^{-34}} 	imes 1.6 	imes 10^{-19} = 1.82 	imes 10^{15}	ext{ Hz}$$

For $n=3 ightarrow n=1$:
$$
u_2 = \frac{E_3 - E_1}{h} = \frac{-6.04 - (-54.4)}{6.63 	imes 10^{-34}} 	imes 1.6 	imes 10^{-19} = 11.67 	imes 10^{15}	ext{ Hz}$$

For $n=2 ightarrow n=1$:
$$
u_3 = \frac{E_2 - E_1}{h} = \frac{-13.6 - (-54.4)}{6.63 	imes 10^{-34}} 	imes 1.6 	imes 10^{-19} = 9.84 	imes 10^{15}	ext{ Hz}$$

And there we have it! By carefully applying the laws of conservation and the principles of quantum mechanics, we've completely unraveled the mysteries of this subatomic collision.

Visualized Solution

$C o l l i s i o n S e t u p$

$C o n ser v a t i o n o f M o m e n t u m$

$M o m e n t u m E q u a t i o n s$

$S o l v in g M o m e n t u m E q u a t i o n s$

$C o n ser v a t i o n o f E n er g y$

$E n er g y L e v e l so f H e^{+}$

$P oss ib l e E x c i t a t i o n E n er g i es$

$C a se 1 : E x c i t a t i o n t o n = 2$

$C a se 2 : E x c i t a t i o n t o n = 3$

$C a se 3 : E x c i t a t i o n t o n = 4$

$F r e q u e n c i eso f E mi tt e d R a d ia t i o n$

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

Setting the Stage

The Math of Momentum

The Energy Equation

Exploring the Quantum Realm

The Moment of Truth

The Quantum Limit

The Aftermath

Similar Questions

A free electron of $2.6 eV$ energy collides with a $H^{+}$ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. ( $h = 6.6 \times 1 0^{- 34} Js$ )

Comprehension Passage

The quantum number $n$ of the state finally populated in He $^{+}$ ions is

The wavelength of light emitted in the visible region by He $^{+}$ ions after collisions with H atoms is

The ratio of the kinetic energy of the $n = 2$ electron for the H atom to that of He $^{+}$ ion is

An excited $He^{+}$ ion emits two photons in succession, with wavelengths $108.5 nm$ and $30.4 nm$ , in making a transition to ground state. The quantum number $n$ corresponding to its initial excited state is [for photon of wavelength $λ$ , energy $E = \frac{1240 eV}{λ (in nm)}$ ]

Radiation coming from transitions $n = 2$ to $n = 1$ of hydrogen atoms fall on $He^{+}$ ions in $n = 1$ and $n = 2$ states. The possible transition of helium ions as they absorb energy from the radiation is

Visualized Solution

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