A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electro

Q: Why is the final kinetic energy zero?

At the distance of closest approach, the proton momentarily comes to rest before reversing its direction due to the repulsive electrostatic force. Hence, its kinetic energy is zero at that instant.

Q: Why do we ignore the recoil of the nucleus?

The nucleus is much heavier than the proton (it has a charge of $120e$, implying a very large mass number). Therefore, its recoil velocity and kinetic energy are negligible compared to the proton.

Q: Can we use classical mechanics for this problem?

Yes, for the trajectory and energy conservation, classical mechanics is a very good approximation here. We only use quantum mechanics at the end to find the de-Broglie wavelength.

Q: What happens if the proton is fired with more energy?

If the proton has more initial kinetic energy, it will overcome the electrostatic repulsion for a longer distance, resulting in a smaller distance of closest approach.

Q: Why is the answer rounded to 7?

The exact calculation gives approximately $6.9 \text{ fm}$. Since the question asks for an integer answer, we round it to the nearest integer, which is 7.

Question

A proton is fired from very far away towards a nucleus with charge $Q = 120 e$, where $e$ is the electronic charge. It makes a closest approach of $10 	ext{ fm}$ to the nucleus. The de-Broglie wavelength (in units of fm) of the proton at its start is [Take the proton mass, $m_p = (5/3) 	imes 10^{-27} 	ext{ kg}$; $h = 6.63 	imes 10^{-34} 	ext{ J-s}$; $e = 1.6 	imes 10^{-19} 	ext{ C}$; $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 	ext{ m/F}$; $1 	ext{ fm} = 10^{-15} 	ext{ m}$]

Sigma Percentile · Accepted Answer

### The Setup: A Subatomic Game of Billiards

Imagine a game of billiards, but on a subatomic scale. A proton is fired from a vast distance towards a massive nucleus. As the proton approaches, it doesn't just crash into the nucleus. Instead, it encounters a powerful invisible force field—the electrostatic repulsion between the positively charged proton and the positively charged nucleus.

This repulsive force acts like a stiff spring, gradually slowing the proton down. Eventually, the proton comes to a momentary halt before being pushed back. The point where it stops is called the **distance of closest approach**, denoted by $r$.

### Energy Conservation: The Invisible Spring

To solve this problem, we rely on one of the most fundamental principles in physics: the conservation of mechanical energy.

Initially, the proton is so far away that its electrostatic potential energy is effectively zero. All its energy is purely kinetic, which we'll call $K$.

As it reaches the distance of closest approach, its velocity drops to zero, meaning its kinetic energy vanishes. Where did all that energy go? It has been entirely converted into electrostatic potential energy!

We can write this mathematically as:
$$K_i + U_i = K_f + U_f$$

Substituting our known states:
$$K + 0 = 0 + \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}$$

### Calculating the Kinetic Energy

Now, let's plug in the specific charges given in the problem. The nucleus has a massive charge of $Q = 120e$, and our projectile is a single proton with charge $e$. The distance of closest approach is given as $r = 10 	ext{ fm}$.

Substituting these into our energy equation gives us the initial kinetic energy of the proton:
$$K = \frac{1}{4\pi\varepsilon_0} \frac{(120e)(e)}{r}$$

### The Quantum Connection: de-Broglie Wavelength

The problem doesn't just ask for the energy; it asks for the **de-Broglie wavelength** of the proton at the start. This is where classical mechanics meets quantum mechanics.

Louis de Broglie proposed that every moving particle has an associated wave, with a wavelength $\lambda$ given by:
$$\lambda = \frac{h}{p}$$

Since momentum $p$ is related to kinetic energy by $p = \sqrt{2m_p K}$, we can rewrite the wavelength formula as:
$$\lambda = \frac{h}{\sqrt{2m_p K}}$$

### Crunching the Numbers

This is the part where we need to be incredibly careful with our powers of 10. Let's calculate the momentum $p$ first:
$$p = \sqrt{2m_p \left( \frac{1}{4\pi\varepsilon_0} \frac{120e^2}{r} ight)}$$

Plugging in the constants:
$$p = \sqrt{2 \left(\frac{5}{3} 	imes 10^{-27}ight) \left(9 	imes 10^9ight) \frac{120(1.6 	imes 10^{-19})^2}{10 	imes 10^{-15}}}$$

Let's simplify the terms inside the square root:
$$p = \sqrt{\frac{10}{3} 	imes 10^{-27} 	imes 9 	imes 10^9 	imes 12 	imes 10^{15} 	imes 2.56 	imes 10^{-38}}$$
$$p = \sqrt{30 	imes 10^{-18} 	imes 12 	imes 10^{15} 	imes 2.56 	imes 10^{-38}}$$
$$p = \sqrt{360 	imes 10^{-3} 	imes 2.56 	imes 10^{-38}}$$
$$p = \sqrt{921.6 	imes 10^{-41}} = \sqrt{9216 	imes 10^{-42}}$$

Taking the square root gives us a clean, manageable number:
$$p = 96 	imes 10^{-21} 	ext{ kg m/s}$$

### The Final Result

With the momentum calculated, finding the de-Broglie wavelength is just one division away:
$$\lambda = \frac{6.63 	imes 10^{-34}}{96 	imes 10^{-21}}$$
$$\lambda = 0.06906 	imes 10^{-13} 	ext{ m}$$
$$\lambda = 6.9 	imes 10^{-15} 	ext{ m}$$

Since $1 	ext{ fm} = 10^{-15} 	ext{ m}$, the wavelength is $6.9 	ext{ fm}$.

The question asks for the answer as an integer, so we round $6.9$ to the nearest whole number.

**Final Answer: 7**

Enter Numerical Value:

Visualized Solution

$r = 10 fm$

$K_{i} + U_{i} = K_{f} + U_{f}$

$K = U_{f}$

$K = \frac{1}{4 π ε _{0}} \frac{( 120 e ) ( e )}{r}$

$λ = \frac{h}{2 m _{p} K}$

$p = 96 \times 1 0^{- 21} kg m/s$

$λ \approx 7 fm$

$What if α particle?$

The Sigma Insight: Matter Waves and de Broglie Relation

The Setup

Energy Conservation

Calculating the Kinetic Energy

The Quantum Connection: de-Broglie Wavelength

Crunching the Numbers

The Final Result

Similar Questions

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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; } r=10 fm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ki​+Ui​=Kf​+Uf​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=4πε0​1​r(120e)(e)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=2mp​K​h​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=96×10−21 kg m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ≈7 fm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if α particle?