The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of

Q: Why do we use $E = \frac{3}{2}k_B T$ for an electron?

In thermal equilibrium, the electron is treated as a monatomic gas particle moving in 3-dimensional space. Therefore, it has 3 degrees of freedom, and by the equipartition theorem, its average translational kinetic energy is $\frac{3}{2}k_B T$.

Q: Can we just use $E = k_B T$?

No, $E = k_B T$ would correspond to a system with only 2 degrees of freedom. For a particle free to move in all three spatial dimensions, you must use $\frac{3}{2}k_B T$.

Q: Why is the electron considered part of an ideal gas?

It's a standard physics approximation. When an electron is in thermal equilibrium with a gas at temperature $T$, it acquires the same average thermal kinetic energy as the gas molecules, behaving like an ideal gas particle itself.

Q: What happens to the de Broglie wavelength if the temperature increases?

As temperature increases, the thermal kinetic energy and momentum of the electron increase. Since wavelength is inversely proportional to momentum ($\lambda \propto \frac{1}{\sqrt{T}}$), the de Broglie wavelength will decrease.

Q: Why are the powers of 10 so critical in this calculation?

When taking the square root of a number with an exponent, the exponent must be even. A single miscounted zero can lead to an odd exponent, forcing you to shift decimal places and potentially throwing your final answer off by a massive factor. Always combine and verify your powers of 10 before applying the square root!

Question

The temperature of an ideal gas in 3-dimensions is $300	ext{ K}$. The corresponding de-Broglie wavelength of the electron approximately at $300	ext{ K}$, is
[$m_e = 	ext{mass of electron} = 9 	imes 10^{-31}	ext{ kg}$
$h = 	ext{Planck's constant} = 6.6 	imes 10^{-34}	ext{ Js}$
$K_B = 	ext{Boltzmann constant} = 1.38 	imes 10^{-23}	ext{ JK}^{-1}$]

Options:

(A) $6.26	ext{ nm}$

(B) $8.46	ext{ nm}$

(C) $2.26	ext{ nm}$

(D) $3.25	ext{ nm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine an electron moving around inside a three-dimensional ideal gas. The temperature of the gas is $300	ext{ K}$. Because of this temperature, the electron possesses thermal kinetic energy.

According to the de Broglie hypothesis, any moving particle has an associated wave. We need to find the exact wavelength of this wave for our thermal electron.

### The Master Equation
To find the wavelength, we use the fundamental de Broglie formula:
$$ \lambda = \frac{h}{p} $$

Here, $p$ is the momentum of the electron. We can express the momentum in terms of kinetic energy $E$ as $p = \sqrt{2mE}$. Substituting this, we get:
$$ \lambda = \frac{h}{\sqrt{2mE}} $$

Now, what is the kinetic energy of this electron? From the kinetic theory of gases, we know that the average translational kinetic energy in a three-dimensional gas is given by:
$$ E = \frac{3}{2}k_B T $$

Let's substitute this kinetic energy value into our wavelength formula. The factor of $2$ in the numerator and denominator will cancel out beautifully. 
$$ \lambda = \frac{h}{\sqrt{2m \left(\frac{3}{2}k_B Tight)}} $$

This simplifies to our master equation:
$$ \lambda = \frac{h}{\sqrt{3mk_B T}} $$

### Final Calculation
Let's substitute the given values into our master equation and get the answer. Carefully place the values of Planck's constant, mass, Boltzmann constant, and temperature.
$$ \lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{3 	imes 9 	imes 10^{-31} 	imes 1.38 	imes 10^{-23} 	imes 300}} $$

Let's multiply the terms inside the square root. $3 	imes 9 	imes 300 	imes 1.38$ gives $11178$. And combining the powers of ten gives $10^{-54}$.
$$ \lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{11178 	imes 10^{-54}}} $$

The square root of $11178$ is approximately $105.72$. Dividing $6.6$ by this value gives us:
$$ \lambda = \frac{6.6 	imes 10^{-34}}{105.72 	imes 10^{-27}} \approx 0.0624 	imes 10^{-7}	ext{ m} $$

Which we can write approximately as **$6.26	ext{ nm}$**. And that is our final answer!

Select Answer:

Visualized Solution

\text{Visualizing the Thermal Electron}

\text{de-Broglie Wavelength Formula}

\text{Thermal Kinetic Energy}

\text{The Master Equation}

\text{Substituting the Values}

\text{Simplifying the Expression}

\text{Final Calculation}

The Sigma Insight: Matter Waves and de Broglie Relation

Analyzing the Setup Imagine an electron moving around inside a three-dimensional ideal gas

The Master Equation

Final Calculation Let's substitute the given values into our master equation and get the answer

Similar Questions

An electron (of mass $m$ ) and a photon have the same energy $E$ in the range of a few electron volt. The ratio of the de Broglie wavelength associated with the electron and the wavelength of the photon is ( $c =$ speed of light in vacuum)

An electron of mass $m$ and a photon have same energy $E$ . The ratio of wavelength of electron to that of photon is ( $c$ being the velocity of light)

If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor

An electron (mass $m$ ) with initial velocity $v = v_{0} \hat{i} + v_{0} \hat{j}$ is in an electric field $E = - E_{0} \hat{k}$ . If $λ_{0}$ is initial de-Broglie wavelength of electron, then its de Broglie wavelength at time $t$ is given by

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

Electrons with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

A particle moving with kinetic energy $E$ has de Broglie wavelength $λ$ . If energy $Δ E$ is added to its energy, the wavelength become $λ /2$ . Value of $Δ E$ is

Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to

An electron moving with speed $v$ and a photon moving with speed $c$ , have same de-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is