Question

Sea of electrons, which are free electrons in conductor lattice, can be called as Fermi gas. At temperature $T$, the de Broglie wavelength of electrons is $\lambda$. If temperature of conductor is increased by $\Delta T (\Delta T \ll T)$, then new de Broglie wavelength of electron is best given by

• A. $\lambda \left(1+\frac{\Delta T}{T}\right)$
• B. $\lambda \left(1-\frac{\Delta T}{T}\right)$
• C. $\lambda \left(1-\frac{\Delta T}{2T}\right)$
• D. $\lambda \left(1+\frac{\Delta T}{2T}\right)$

Answer 1

Answer: (C)

$\lambda \left(1-\frac{\Delta T}{2T}\right)$

Answer 2

The de Broglie wavelength ($\lambda$) of a particle is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum. The momentum is related to the kinetic energy ($E_k$) by $p = \sqrt{2mE_k}$, where $m$ is the mass of the particle. Thus, the de Broglie wavelength is inversely proportional to the square root of the kinetic energy: $\lambda \propto \frac{1}{\sqrt{E_k}}$ For a classical gas, the average kinetic energy per particle is directly proportional to the absolute temperature: $E_k \propto T$. Although the question mentions "Fermi gas," the options provided suggest a simplified model where the temperature dependence of the relevant electron energy follows the classical gas behavior. Therefore, we can assume: $E_k \propto T$ Combining these relationships, we get: $\lambda \propto \frac{1}{\sqrt{T}}$ Let $\lambda_1$ be the de Broglie wavelength at temperature $T_1 = T$, and $\lambda_2$ be the new wavelength at temperature $T_2 = T + \Delta T$. We can write: $\lambda_1 = C \cdot T^{-1/2}$ $\lambda_2 = C \cdot (T + \Delta T)^{-1/2}$ where $C$ is a constant of proportionality. Now, we can express $\lambda_2$ in terms of $\lambda_1$: $\lambda_2 = C \cdot T^{-1/2} \left(1 + \frac{\Delta T}{T}\right)^{-1/2}$ $\lambda_2 = \lambda_1 \left(1 + \frac{\Delta T}{T}\right)^{-1/2}$ Given that $\Delta T \ll T$, we can use the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$. Here, $x = \frac{\Delta T}{T}$ and $n = -\frac{1}{2}$. $\left(1 + \frac{\Delta T}{T}\right)^{-1/2} \approx 1 + \left(-\frac{1}{2}\right) \left(\frac{\Delta T}{T}\right) = 1 - \frac{\Delta T}{2T}$ Substituting this approximation back into the expression for $\lambda_2$: $\lambda_2 \approx \lambda_1 \left(1 - \frac{\Delta T}{2T}\right)$ Thus, the new de Broglie wavelength is approximately $\lambda \left(1 - \frac{\Delta T}{2T}\right)$.