de Broglie wavelength $\lambda$ as a function of $\dfrac{1}{\sqrt{K}}$, for two particles of masses $m_1$ and $m_2$ are shown in the figure. Here, $K$ is the energy of the moving particles. (a) What does the slope of a line represent ? (b) Which of the two particles is heavier ?

Question

de Broglie wavelength $\lambda$ as a function of $\dfrac{1}{\sqrt{K}}$, for two particles of masses $m_1$ and $m_2$ are shown in the figure. Here, $K$ is the energy of the moving particles. (a) What does the slope of a line represent ? (b) Which of the two particles is heavier ? (c) Is this graph also valid for a photon ? Justify your answer in each case.

![Figure: lambda vs 1/sqrt(K) graph](/figures/pyq/2024-55-2-1/fig-q27.svg)

Accepted Answer

> [!TLDR]
> The de Broglie wavelength $\lambda$ relates to kinetic energy $K$ as $\lambda = h/\sqrt{2mK}$, so a plot of $\lambda$ vs $1/\sqrt{K}$ is a straight line through the origin whose slope is $h/\sqrt{2m}$. The steeper line corresponds to the lighter particle, so $m_1$ is heavier. This graph does **not** apply to photons because the relation $p = \sqrt{2mK}$ is invalid for massless particles.

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The core idea here is the de Broglie relation $\lambda = h/p$, combined with the classical expression for kinetic energy $K = p^2/2m$. When you eliminate momentum $p$, you get $\lambda = h/\sqrt{2mK}$. This is a power-law relation: $\lambda \propto 1/\sqrt{K}$, with the constant of proportionality depending on mass. Plotting $\lambda$ against $1/\sqrt{K}$ therefore gives a straight line through the origin — the slope carries the mass information.

Let’s work through each part systematically.

1. **Finding what the slope represents**  
   Start from the de Broglie wavelength for a massive particle:  
   $$ \lambda = \frac{h}{p} $$  
   For a non-relativistic particle, kinetic energy $K = p^2/2m$, so $p = \sqrt{2mK}$. Substituting:  
   $$ \lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2m}} \cdot \frac{1}{\sqrt{K}} $$  
   This is of the form $y = (	ext{slope}) \cdot x$, where $y = \lambda$ and $x = 1/\sqrt{K}$.  
   Therefore, the slope of each line is:  
   $$ 	ext{slope} = \frac{h}{\sqrt{2m}} $$  
   The slope depends only on the particle’s mass (and Planck’s constant). A heavier particle gives a smaller slope; a lighter particle gives a larger slope.

2. **Which particle is heavier?**  
   From the graph, the line labelled $m_2$ is steeper than the line labelled $m_1$.  
   Since slope $\propto 1/\sqrt{m}$, a steeper slope means a smaller mass.  
   So $m_2$ is the lighter particle, and $m_1$ is the heavier particle.

> [!WARNING]  
   > A common mistake is to think “steeper line = heavier particle” because a steeper line rises faster. But here the slope is

Q.de Broglie wavelength $\lambda$ as a function of $\dfrac{1}{\sqrt{K}}$ , for two particles of masses $m_1$ and $m_2$ are shown in the figure. Here, $K$ is the energy of the moving particles.

Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

The Core Idea

Why You Don't See It in Daily Life

What This Means Physically

Why this formula?

De Broglie Wavelength: Why the Formula Holds

The Core Insight: Nature's Symmetry

Step 1: Start with Light (What We Already Knew)

Step 2: De Broglie's Bold Hypothesis

Step 3: Why Momentum and Not Velocity?

Step 4: The Deeper Reasoning — Wave-Particle Duality

Step 5: Experimental Confirmation (Why We Believe It)

Q.de Broglie wavelength λ\lambdaλ as a function of 1K\dfrac{1}{\sqrt{K}}K​1​, for two particles of masses m1m_1m1​ and m2m_2m2​ are shown in the figure. Here, KKK is the energy of the moving particles.

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Q.de Broglie wavelength $\lambda$ as a function of $\dfrac{1}{\sqrt{K}}$ , for two particles of masses $m_1$ and $m_2$ are shown in the figure. Here, $K$ is the energy of the moving particles.

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