A spectrometer is used to detect plasma oscillations in a sample. The spectrometer
can work in the range of 3 × 1012 rad s-1 to 30 × 1012 rad s-1. The minimum
carrier concentration that can be detected by using this spectrometer is n × 1021
m-. The value of n is _____________ (Round off to two decimal places)
(Charge of an electron = -1.6 × 10-1 C, mass of an electron = 9.1 × 10-^31 kg
and \epsilon_0 = 8.85 × 10-12 C2 N-1 m-2)
Official Answer Key:
2.70 to 2.96
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.35)**
**1. Solid State Physics Principles (Plasma Oscillations)**:
- The electron plasma frequency $\omega_p$ in a conductor or semiconductor is:
$\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$
$\implies n_e = \frac{\epsilon_0 m_e \omega_p^2}{e^2}$
**2. Detailed Step-by-Step Derivation**:
- To detect plasma oscillations within the range $3 \times 10^{12} \le \omega \le 30 \times 10^{12} \text{ rad s}^{-1}$, the minimum carrier concentration corresponds to the lower limit:
$\omega_{p, \text{min}} = 3 \times 10^{12} \text{ rad s}^{-1}$
- Substitute the given physical constants:
- $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$
- $m_e = 9.1 \times 10^{-31} \text{ kg}$
- $e = 1.6 \times 10^{-19} \text{ C} \implies e^2 = 2.56 \times 10^{-38} \text{ C}^2$
- $\omega_{p, \text{min}}^2 = (3 \times 10^{12})^2 = 9 \times 10^{24} \text{ rad}^2\text{ s}^{-2}$
- Compute $n_{e, \text{min}}$:
$n_{e, \text{min}} = \frac{(8.85 \times 10^{-12}) \times (9.1 \times 10^{-31}) \times (9 \times 10^{24})}{2.56 \times 10^{-38}}$
$n_{e, \text{min}} = \frac{724.815 \times 10^{-18}}{2.56 \times 10^{-38}} = 283.13 \times 10^{20} = 2.831 \times 10^{21} \text{ m}^{-3}$
- Given in the form $n \times 10^{21} \text{ m}^{-3}$, the coefficient is $n \approx 2.83$.
**3. Official Verification**:
The verified official IIT Master Answer range is **2.70 to 2.96**.
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