An electron, a doubly ionised helium ion (He++) and a proton are having the same kinetic energy. The

Q: Why do we use the kinetic energy formula instead of the voltage formula?

The question explicitly states that the particles have the 'same kinetic energy'. If the question had stated they were accelerated through the 'same potential difference', we would use the formula $\lambda = \frac{h}{\sqrt{2mqV}}$, where charge ($q$) would also play a role.

Q: What exactly is a doubly ionized helium ion?

A normal helium atom has 2 protons, 2 neutrons, and 2 electrons. When it is 'doubly ionized', it loses both of its electrons. What remains is just the nucleus (2 protons and 2 neutrons), which is also known as an alpha particle. Its mass is roughly 4 times that of a single proton.

Q: How much heavier is a proton compared to an electron?

A proton is approximately 1836 times heavier than an electron. This massive difference is why the electron's wavelength will be significantly larger than the proton's when they share the same kinetic energy.

Q: Does the charge of the particles matter in this specific question?

No, the charge does not matter here. The de-Broglie wavelength formula $\lambda = \frac{h}{\sqrt{2mK}}$ only depends on mass and kinetic energy. Charge only becomes relevant if the kinetic energy is acquired by accelerating the particle through an electric potential ($K = qV$).

Q: What if the particles had the same momentum instead of kinetic energy?

If they had the same momentum ($p$), then according to the fundamental de-Broglie relation $\lambda = \frac{h}{p}$, all three particles would have the exact same wavelength, regardless of their drastically different masses!

Question

An electron, a doubly ionised helium ion ($\mathrm{He}^{++}$) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_e$, $\lambda_{\mathrm{He}^{++}}$ and $\lambda_p$ is Options: (A) $\lambda_e > \lambda_{\mathrm{He}^{++}} > \lambda_p$ (B) $\lambda_e < \lambda_{\mathrm{He}^{++}} = \lambda_p$ (C) $\lambda_e > \lambda_p > \lambda_{\mathrm{He}^{++}}$ (D) $\lambda_e < \lambda_p < \lambda_{\mathrm{He}^{++}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine a microscopic race track where three distinct particles are zooming past: a tiny electron ($e^-$), a much heavier proton ($p$), and a massive doubly ionized helium ion ($\mathrm{He}^{++}$), which is essentially an alpha particle. The problem gives us a very specific constraint: all three particles possess the exact same **kinetic energy ($K$)**. Our mission is to figure out how their de-Broglie wavelengths compare to one another. ### The Master Equation To solve this, we need a bridge that connects the wave nature of a particle (its wavelength) to its particle nature (its kinetic energy). Enter the de-Broglie wavelength formula: $$\lambda = \frac{h}{p}$$ Here, $h$ is Planck's constant and $p$ is the momentum of the particle. But we are given kinetic energy, not momentum. We know from classical mechanics that kinetic energy $K = \frac{p^2}{2m}$. Rearranging this for momentum gives us $p = \sqrt{2mK}$. Substituting this back into the de-Broglie equation, we get our master equation for this problem: $$\lambda = \frac{h}{\sqrt{2mK}}$$ ### The Proportionality Logic Look closely at the master equation. The problem states that the kinetic energy ($K$) is the same for all three particles. Planck's constant ($h$) is, of course, a universal constant, and $2$ is just a number. Since $h$ and $K$ are constant, the only variable dictating the wavelength is the mass ($m$) of the particle. We can extract a beautiful, simple proportionality: $$\lambda \propto \frac{1}{\sqrt{m}}$$ This tells us a profound physical truth: for particles with the same kinetic energy, the heavier the particle, the shorter its de-Broglie wavelength. ### Final Calculation Now, all we have to do is rank the particles by their mass. 1. The **electron** is the lightest fundamental particle among the three ($m_e$). 2. The **proton** is significantly heavier, roughly $1836$ times the mass of an electron ($m_p \approx 1836 m_e$). 3. The **doubly ionized helium ion** (alpha particle) consists of two protons and two neutrons, making it about $4$ times heavier than a single proton ($m_{\mathrm{He}^{++}} \approx 4 m_p$). So, the order of their masses is strictly: $$m_e < m_p < m_{\mathrm{He}^{++}}$$ Because the wavelength is inversely proportional to the square root of the mass, the order of their wavelengths will be the exact reverse of their mass order. The lightest particle gets the longest wave, and the heaviest gets the shortest: $$\lambda_e > \lambda_p > \lambda_{\mathrm{He}^{++}}$$ This perfectly matches option (c).

Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron, a doubly ionised helium ion ( $He^{++}$ ) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $λ_{e}$ , $λ_{He^{++}}$ and $λ_{p}$ is

Select Answer:

Visualized Solution

Visualizing the Particles

de-Broglie Wavelength Formula

Establishing Proportionality

Comparing Masses

Final Wavelength Order

The Way Forward

The Sigma Insight: Matter Waves and de Broglie Relation

Analyzing the Setup

The Master Equation To solve this, we need a bridge that connects the wave nature of a particle (its wavelength) to its particle nature (its kinetic energy)

The Proportionality Logic Look closely at the master equation

Final Calculation

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