A particle of mass 200 MeV/c^2 collides with a hydrogen atom at rest. Soon after the collision, the

Q: Why is the collision considered inelastic?

In a perfectly elastic collision, the total kinetic energy before and after the collision remains the same. Here, some of the initial kinetic energy is absorbed by the hydrogen atom to excite its electron to a higher energy level. Because macroscopic kinetic energy is converted into internal quantum energy, the collision is inelastic.

Q: How did we get the mass of the hydrogen atom as $5m$?

The mass of the particle is given as $200 \text{ MeV/c}^2$. The mass of the hydrogen atom is given as $1 \text{ GeV/c}^2$. Since $1 \text{ GeV} = 1000 \text{ MeV}$, the hydrogen atom's mass is $1000 \text{ MeV/c}^2$. Dividing $1000$ by $200$ gives exactly $5$. So, if the particle's mass is $m$, the hydrogen atom's mass is $5m$.

Q: Why is the final velocity of the incoming particle zero?

This is explicitly stated in the problem description: 'Soon after the collision, the particle comes to rest'. We must strictly follow the conditions given in the question.

Q: What does 'first excited state' mean? Is it $n=1$?

No, the ground state (the lowest energy level) is $n=1$. The 'first excited state' is the next level up, which is $n=2$. The 'second excited state' would be $n=3$, and so on.

Q: Why do we use $13.6 \text{ eV}$ in the energy calculation?

$13.6 \text{ eV}$ is the standard Rydberg energy for a hydrogen atom. The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by the formula $E_n = -13.6 / n^2 \text{ eV}$. This is a fundamental constant you should memorize for JEE.

Question

A particle of mass $200 	ext{ MeV/c}^2$ collides with a hydrogen atom at rest. Soon after the collision, the particle comes to rest and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is $N / 4$. The value of $N$ is ......... .
(Given, the mass of the hydrogen atom to be $1 	ext{ GeV/c}^2$)

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of classical mechanics and quantum physics. We are dealing with an inelastic collision where macroscopic kinetic energy is converted into the internal quantum excitation energy of an atom. Let's break down the thought process step by step.

### Analyzing the Setup

Imagine a particle with a mass of $200 	ext{ MeV/c}^2$ hurtling towards a stationary hydrogen atom. The mass of the hydrogen atom is given as $1 	ext{ GeV/c}^2$.

Before we do any physics, we must align our units. Since $1 	ext{ GeV} = 1000 	ext{ MeV}$, the mass of the hydrogen atom is $1000 	ext{ MeV/c}^2$. Notice the elegant relationship here: the hydrogen atom is exactly $5$ times more massive than the incoming particle.

Let the mass of the particle be $m$. Therefore, the mass of the hydrogen atom is $M = 5m$.

### The Master Equation: Momentum Conservation

During the collision, there are no external forces acting on the system. This means we can safely apply the **Conservation of Linear Momentum**.

The problem states that after the collision, the incoming particle comes to a complete rest, and the hydrogen atom recoils. Let the initial velocity of the particle be $v$, and the recoil velocity of the hydrogen atom be $v'$.

$$p_{	ext{initial}} = p_{	ext{final}}$$

$$mv + 0 = 0 + Mv'$$

Substituting $M = 5m$, we get:

$$mv = 5m v' \implies v' = \frac{v}{5}$$

So, the hydrogen atom recoils with one-fifth of the particle's initial velocity.

### The Energy Balance

This is an inelastic collision. The kinetic energy of the system is not conserved. Where does the lost kinetic energy go? It is absorbed by the hydrogen atom, causing its electron to jump to a higher energy level—specifically, the **first excited state** ($n=2$).

Let's calculate the loss in kinetic energy ($\Delta K$). The initial kinetic energy is just the energy of the incoming particle, which we will call $K$:

$$K = \frac{1}{2}mv^2$$

The final kinetic energy is the energy of the recoiling hydrogen atom:

$$K_{	ext{final}} = \frac{1}{2}Mv'^2 = \frac{1}{2}(5m)\left(\frac{v}{5}ight)^2$$

Simplifying this expression:

$$K_{	ext{final}} = \frac{1}{2}(5m)\left(\frac{v^2}{25}ight) = \frac{1}{5}\left(\frac{1}{2}mv^2ight) = \frac{K}{5}$$

The loss in kinetic energy is the difference between the initial and final kinetic energies:

$$\Delta K = K - \frac{K}{5} = \frac{4}{5}K$$

### Final Calculation

This lost kinetic energy $\frac{4}{5}K$ is exactly equal to the excitation energy required to take the hydrogen atom from its ground state ($n=1$) to its first excited state ($n=2$).

The energy of the $n^{	ext{th}}$ state of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.

The excitation energy $\Delta E$ is:

$$\Delta E = E_2 - E_1 = -\frac{13.6}{2^2} - \left(-\frac{13.6}{1^2}ight) = -3.4 + 13.6 = 10.2 	ext{ eV}$$

Now, we equate the kinetic energy loss to the excitation energy:

$$\frac{4}{5}K = 10.2 	ext{ eV}$$

Solving for $K$:

$$K = \frac{10.2 	imes 5}{4} = \frac{51}{4} 	ext{ eV}$$

The problem states that the initial kinetic energy is $\frac{N}{4} 	ext{ eV}$. Comparing this with our result:

$$\frac{N}{4} = \frac{51}{4} \implies \mathbf{N = 51}$$

And there we have it! A perfect blend of momentum conservation and Bohr's atomic model.

Enter Numerical Value:

Visualized Solution

$Collision Setup$

$Conservation of Linear Momentum$

$Velocity of Recoiling Atom$

$Conservation of Energy$

$Loss in Kinetic Energy$

$Simplifying Δ K$

$Excitation Energy of H-atom$

$Equating Energy Loss to Excitation$

$Calculating Initial Kinetic Energy K$

$Finding N$

$What if it was He^{+} ion?$

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

Analyzing the Setup

The Master Equation

The Energy Balance

Final Calculation

Similar Questions

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

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.

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

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 $h c = 12500 eV- \overset{˚}{A}$ )

Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it was He+ ion?