An electron of mass m_e and a proton of mass m_p = 1836 m_e are moving with the same speed. The rati

Q: Why do we use $p = mv$ instead of $p = \sqrt{2mK}$?

The choice of formula depends entirely on the given conditions. The problem explicitly states that the particles are moving with the 'same speed'. Therefore, using $p = mv$ allows us to easily cancel out the constant velocity $v$. If the problem had stated they had the 'same kinetic energy', then $p = \sqrt{2mK}$ would be the correct choice.

Q: Does a larger mass always mean a smaller de-Broglie wavelength?

Yes, provided the velocity (or kinetic energy) is kept constant. The de-Broglie wavelength is inversely proportional to momentum ($p = mv$). So, if $v$ is constant, a larger mass $m$ results in a larger momentum, which in turn results in a smaller wavelength.

Q: What is the significance of the number 1836?

It is not a random number! In physics, the mass of a proton is experimentally found to be approximately 1836 times the rest mass of an electron. This is a fundamental constant ratio in nature.

Q: Are relativistic effects considered in this calculation?

No. Unless the problem specifies that the particles are moving at speeds comparable to the speed of light (usually $> 0.1c$), we assume classical mechanics applies, and momentum is simply $p = mv$ without the Lorentz factor.

Q: What if the electron and proton had the exact same momentum instead of speed?

If they had the same momentum $p$, then according to $\lambda = \frac{h}{p}$, their de-Broglie wavelengths would be exactly equal. The ratio $\frac{\lambda_e}{\lambda_p}$ would simply be 1, regardless of their mass difference.

Question

An electron of mass $m_e$ and a proton of mass $m_p = 1836 m_e$ are moving with the same speed. The ratio of their de-Broglie wavelength $\frac{\lambda_{	ext{electron}}}{\lambda_{	ext{proton}}}$ will be

Options:

(A) 1

(B) 1836

(C) $\frac{1}{1836}$

(D) 918

Sigma Percentile · Accepted Answer

### Visualizing the Quantum Race

Imagine a race track where an electron and a proton are sprinting side by side. They are moving with the exact same speed, $v$. However, they are vastly different in size. The proton is a heavyweight champion, tipping the scales at $1836$ times the mass of the nimble electron ($m_p = 1836 m_e$).

In the quantum world, moving particles don't just travel in straight lines like bullets; they exhibit wave-like behavior. This is the core of wave-particle duality. Our goal is to find out how their quantum waves—specifically, their de-Broglie wavelengths—compare to each other.

### The Master Equation: de-Broglie Wavelength

To unlock this comparison, we need the fundamental tool of quantum mechanics: the de-Broglie wavelength formula. The wavelength $\lambda$ of a particle is given by Planck's constant $h$ divided by its momentum $p$.

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

Since momentum is the product of mass and velocity ($p = mv$), we can rewrite this as:

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

### Analyzing the Proportionality

Let's look closely at this equation. We know that Planck's constant $h$ is, well, a constant. The problem also explicitly states that both the electron and the proton are moving with the **same speed** $v$.

Because both $h$ and $v$ are constant in this specific scenario, the wavelength $\lambda$ depends entirely on the mass $m$. Specifically, it is **inversely proportional** to the mass.

$$ \lambda \propto \frac{1}{m} $$

This is a profound realization: the heavier the particle, the smaller its wavelength. The lighter the particle, the more stretched out and pronounced its wave nature becomes.

### The Final Calculation

Now, we want the ratio of the electron's wavelength to the proton's wavelength, $\frac{\lambda_e}{\lambda_p}$. Because of the inverse proportionality we just discovered, this ratio of wavelengths will simply be the inverse ratio of their masses.

$$ \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} $$

We are given that the mass of the proton is $1836$ times the mass of the electron ($m_p = 1836 m_e$). Let's substitute this into our equation:

$$ \frac{\lambda_e}{\lambda_p} = \frac{1836 m_e}{m_e} $$

The mass of the electron ($m_e$) beautifully cancels out from the numerator and the denominator.

$$ \frac{\lambda_e}{\lambda_p} = 1836 $$

And there we have it! The electron's de-Broglie wavelength is $1836$ times longer than that of the proton when they travel at the same speed. This perfectly illustrates why quantum wave effects are so much easier to observe in incredibly light particles like electrons compared to heavier ones.

Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron of mass $m_{e}$ and a proton of mass $m_{p} = 1836 m_{e}$ are moving with the same speed. The ratio of their de-Broglie wavelength $\frac{λ _{electron}}{λ _{proton}}$ will be

Select Answer:

Visualized Solution

Problem Setup

de-Broglie Wavelength

Proportionality

Ratio Calculation

Final Result

Conceptual Takeaway

The Sigma Insight: Matter Waves and de Broglie Relation

Visualizing the Quantum Race

The Master Equation: de-Broglie Wavelength

Analyzing the Proportionality

The Final Calculation

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