A moving proton and electron have the same de-Broglie wavelength. If K and p denote the KE and momen

Q: Why do we use $K = p^2/2m$ instead of $K = \frac{1}{2}mv^2$?

While both formulas are correct, $K = p^2/2m$ is much more direct here because we already established that their momenta ($p$) are equal. Using velocity would require an extra step to find how their velocities compare.

Q: Does a larger mass always mean less kinetic energy?

Not always! It depends on what is kept constant. If momentum is constant (like in this problem), then yes, larger mass means less kinetic energy. But if velocity is constant, a larger mass means *more* kinetic energy.

Q: What exactly is a de-Broglie wavelength?

It's a fundamental concept in quantum mechanics which states that all matter exhibits wave-like properties. The wavelength is inversely proportional to the particle's momentum.

Q: Are the velocities of the proton and electron equal here?

No. Since their momenta ($p = mv$) are equal and the proton has a much larger mass, the proton must be moving much slower than the electron.

Q: Can a macroscopic object like a baseball have a de-Broglie wavelength?

Yes, but because its mass (and therefore momentum) is so incredibly large compared to subatomic particles, its wavelength is unimaginably small—too small to ever be measured or observed!

Question

A moving proton and electron have the same de-Broglie wavelength. If $k$ and $p$ denote the KE and momentum, respectively. Then, choose the correct option.

Options:

(A) $K_p < K_e$ and $p_p = p_e$

(B) $K_p = K_e$ and $p_p = p_e$

(C) $K_p < K_e$ and $p_p < p_e$

(D) $K_p > K_e$ and $p_p = p_e$

Sigma Percentile · Accepted Answer

### The Quantum Setup

Imagine a proton and an electron moving through space. The problem presents us with a fascinating quantum scenario: both particles possess the exact same de-Broglie wavelength.

In the quantum realm, every moving particle has an associated matter wave. The wavelength of this wave, known as the de-Broglie wavelength, is a direct bridge between a particle's wave-like and particle-like properties.

### The Master Equation

To unravel the relationship between their momenta and kinetic energies, we must rely on the fundamental de-Broglie equation:

$$ \lambda = \frac{h}{p} $$

Here, $\lambda$ is the wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle. By rearranging this equation, we can express momentum as:

$$ p = \frac{h}{\lambda} $$

Since the problem explicitly states that the wavelength of the proton ($\lambda_p$) equals the wavelength of the electron ($\lambda_e$), and $h$ is a universal constant, it mathematically guarantees that their momenta must be perfectly identical.

**Therefore, $p_p = p_e$.**

### The Kinetic Energy Connection

Now that we have established that their momenta are equal, we need to compare their kinetic energies. In classical mechanics, kinetic energy ($K$) is typically written as $K = \frac{1}{2}mv^2$. However, when dealing with momentum, a much more powerful and direct form of this equation is:

$$ K = \frac{p^2}{2m} $$

Let's set up a ratio to compare the kinetic energy of the proton ($K_p$) to the kinetic energy of the electron ($K_e$):

$$ \frac{K_p}{K_e} = \frac{\frac{p_p^2}{2m_p}}{\frac{p_e^2}{2m_e}} $$

Since $p_p = p_e$, the momentum terms cancel out beautifully, leaving us with an inverse relationship based purely on mass:

$$ \frac{K_p}{K_e} = \frac{m_e}{m_p} $$

### The Final Verdict

This inverse relationship is the key to the entire problem. We know from fundamental physics that a proton is significantly more massive than an electron—approximately 1836 times heavier!

Because $m_p > m_e$, the fraction $\frac{m_e}{m_p}$ is strictly less than 1.

$$ \frac{K_p}{K_e} < 1 \implies K_p < K_e $$

Even though the proton and electron carry the exact same momentum, the lighter electron must travel much faster to achieve that momentum, resulting in a significantly higher kinetic energy. Thus, the correct conclusion is that $K_p < K_e$ and $p_p = p_e$.

Animated Solution for Physics - Dual Nature of Matter and Radiation: A moving proton and electron have the same de-Broglie wavelength. If $k$ and $p$ denote the KE and momentum, respectively. Then, choose the correct option.

Select Answer:

Visualized Solution

Visualizing the Particles

The de-Broglie Equation

Comparing Momenta

Kinetic Energy Formula

Ratio of Kinetic Energies

Final Conclusion

The Way Forward

The Sigma Insight: Matter Waves and de Broglie Relation

The Quantum Setup

The Master Equation

The Kinetic Energy Connection

The Final Verdict

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