De-broglie's wavelength of an electron gas at 300K is `x A^0` then x is...

Answer

To find the De-Broglie wavelength of an electron gas at 300K, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding De-Broglie's Wavelength**:
The De-Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Relating Momentum to Kinetic Energy**:
The momentum \( p \) of a particle can be expressed in terms of its kinetic energy \( E \):
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the particle.

3. **Kinetic Energy of an Electron Gas**:
For an electron gas at temperature \( T \), the average kinetic energy \( E \) can be given by:
\[ E = \frac{3}{2} kT \]
where \( k \) is the Boltzmann constant.

4. **Substituting Kinetic Energy into the Wavelength Formula**:
Now substituting \( E \) into the momentum formula:
\[ p = \sqrt{2m \left(\frac{3}{2} kT\right)} = \sqrt{3mkT} \]
Therefore, the De-Broglie wavelength becomes:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

5. **Substituting Known Values**:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Boltzmann constant \( k = 1.38 \times 10^{-23} \, \text{J/K} \)
- Temperature \( T = 300 \, \text{K} \)

Plugging in these values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 9.1 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}} \]

6. **Calculating the Denominator**:
First, calculate the term inside the square root:
\[ 3 \times 9.1 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300 \approx 1.0572 \times 10^{-26} \]
Taking the square root:
\[ \sqrt{1.0572 \times 10^{-26}} \approx 1.0277 \times 10^{-13} \]

7. **Final Calculation of Wavelength**:
Now substituting back into the wavelength formula:
\[ \lambda \approx \frac{6.63 \times 10^{-34}}{1.0277 \times 10^{-13}} \approx 6.44 \times 10^{-21} \, \text{m} \]
Converting meters to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 64.4 \, \text{angstroms} \]

8. **Final Result**:
Therefore, the value of \( x \) in the question is:
\[ x \approx 64.4 \]

### Conclusion:
The value of \( x \) is approximately **64**.