Answer :
Answer:
Explanation:To estimate the order of thermal speed of electrons at 27°C, we can use the concept of thermal energy and the Maxwell-Boltzmann distribution of particle speeds. The thermal speed of electrons can be estimated using the following steps:
1. **Convert the temperature to Kelvin:**
The given temperature is 27°C. To convert it to Kelvin, we use:
\[
T = 27 + 273 = 300 \, \text{K}
\]
2. **Use the formula for the root-mean-square (rms) speed of electrons:**
The root-mean-square speed of particles (in this case, electrons) is given by:
\[
v_{\text{rms}} = \sqrt{\frac{3k_B T}{m_e}}
\]
where:
- \( k_B \) is the Boltzmann constant (\( k_B \approx 1.38 \times 10^{-23} \, \text{J/K} \))
- \( T \) is the temperature in Kelvin
- \( m_e \) is the mass of an electron (\( m_e \approx 9.11 \times 10^{-31} \, \text{kg} \))
3. **Substitute the values into the formula:**
\[
v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{9.11 \times 10^{-31}}}
\]
4. **Calculate the expression inside the square root:**
\[
\frac{3 \times 1.38 \times 10^{-23} \times 300}{9.11 \times 10^{-31}} = \frac{1.242 \times 10^{-20}}{9.11 \times 10^{-31}} \approx 1.363 \times 10^{10}
\]
5. **Take the square root of the result:**
\[
v_{\text{rms}} = \sqrt{1.363 \times 10^{10}} \approx 3.69 \times 10^5 \, \text{m/s}
\]
So, the thermal speed (root-mean-square speed) of electrons at 27°C is approximately \( 3.69 \times 10^5 \) m/s. This gives an order of magnitude of about \( 10^5 \) m/s.
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The thermal speed of electrons can be calculated using the formula:
v = sqrt(3kT/m)
Where:
v = thermal speed
k = Boltzmann constant (1.38 x 10^-23 J/K)
T = temperature in Kelvin (27°C = 300 K)
m = mass of an electron (9.11 x 10^-31 kg)
Plugging in the values:
v = sqrt(3 * 1.38 x 10^-23 J/K * 300 K / 9.11 x 10^-31 kg)
v = sqrt(1.24 x 10^-21 m^2/s^2)
v ≈ 1.11 x 10^10 m/s
Therefore, the order of thermal speed of electrons at 27°C is approximately 10^10 m/s.
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