Animated Solution for Physics - Dual Nature of Matter and Radiation: Assuming the nitrogen molecule is moving with rms velocity at $400\text{ K}$, the de-Broglie wavelength of nitrogen molecule is close to (Given, weight of ${\text{N}}_2$ molecule $=4.64\times 10^{-26}\text{ kg}$, Boltzmann constant $=1.38\times 10^{-23}\text{ J/K}$ and Planck's constant $=6.63\times 10^{-34}\text{ J-s}$)

Imagine a nitrogen molecule zipping around randomly in a container at $400\text{ K}$. Classical physics tells us it's just a tiny particle bouncing off walls. But quantum mechanics, specifically Louis de Broglie's brilliant hypothesis, reveals a deeper truth: every moving particle has a wave associated with it.

This isn't just a mathematical trick; it's the fundamental nature of reality. The faster the particle moves, or the heavier it is, the shorter its wavelength. Today, we are going to calculate the exact wavelength of this microscopic quantum dance.

To find this wavelength, we start with the legendary de Broglie equation:

Here, $h$ is Planck's constant and $p$ is the momentum of the molecule. But we aren't given the momentum directly; we are given the temperature $T$. This is where the Kinetic Theory of Gases bridges the gap. For an ideal gas molecule, its average translational kinetic energy $K$ is directly proportional to its absolute temperature:

We also know the classic relationship between kinetic energy and momentum: $p=\sqrt{2mK}$. By substituting our expression for $K$ into this momentum equation, we get:

Now, we substitute this momentum back into our de Broglie equation to forge our master formula:

This is where many students make silly mistakes. We have a beautiful formula, but the numbers are messy. Let's carefully substitute the given values:

Let's tackle the denominator first. Ignore the powers of ten for a moment and multiply the coefficients: $3\times 1.38\times 400\times 4.64=7683.84$.

Next, combine the powers of ten: $10^{-23}\times 10^{-26}=10^{-49}$. So, the term inside the square root is $7683.84\times 10^{-49}$.

Here is a pro-tip: Never try to take the square root of an odd power of ten. Shift the decimal point to make the exponent even. Let's rewrite it as $768.384\times 10^{-48}$.

Now, we need the square root of $768.384$. We know $27^2=729$ and $28^2=784$. Since $768$ is closer to $784$, the square root will be roughly $27.7$. The square root of $10^{-48}$ is simply $10^{-24}$.

Dividing $6.63$ by $27.72$ gives approximately $0.239$. Subtracting the exponents ($-34-(-24)$) leaves us with $10^{-10}$.

Since $10^{-10}\text{ m}$ is exactly $1\text{ A˚}$ (Angstrom), we can elegantly write our final answer as:

And there you have it! The quantum wavelength of a nitrogen molecule at room temperature is roughly a quarter of an Angstrom—smaller than the size of an atom itself. Keep practicing these calculations, and the math will become second nature!