The de Broglie wavelength of an electron in a metal at `27^(@)C` is
`("Given"m_(e)=9.1xx10^(-31)kg,k_(B)=1.38xx10^(-23)JK^(-1))`

Options

A. `6.2xx10^(-9)m`
B. `6.2xx10^(-10)m`
C. `6.2xx10^(-8)m`
D. `6.2xx10^(-7)m`

Explanation

To find the de Broglie wavelength of an electron in a metal at \(27^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the electron.

2. **Relate Momentum to Energy**:
The momentum \(p\) can be expressed in terms of kinetic energy \(E\) as follows:
\[ p = \sqrt{2mE} \]
where \(m\) is the mass of the electron.

3. **Use the Thermal Energy Formula**:
For an electron in a metal at a given temperature, the kinetic energy can be approximated using the formula:
\[ E = \frac{3}{2} k_B T \]
where \(k_B\) is the Boltzmann constant and \(T\) is the absolute temperature in Kelvin.

4. **Convert Temperature to Kelvin**:
The given temperature is \(27^\circ C\). To convert this to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

5. **Substitute Energy into Momentum**:
Substitute \(E\) into the momentum equation:
\[ p = \sqrt{2m \left(\frac{3}{2} k_B T\right)} = \sqrt{3m k_B T} \]

6. **Substitute Momentum into the Wavelength Formula**:
Now substitute \(p\) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{3m k_B T}} \]

7. **Insert Known Values**:
Use the following values:
- \(h = 6.63 \times 10^{-34} \, J \cdot s\)
- \(m_e = 9.1 \times 10^{-31} \, kg\)
- \(k_B = 1.38 \times 10^{-23} \, J/K\)
- \(T = 300 \, K\)

Now, substituting these values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times (9.1 \times 10^{-31}) \times (1.38 \times 10^{-23}) \times 300}} \]

8. **Calculate the Denominator**:
First, calculate the value inside the square root:
\[ 3 \times (9.1 \times 10^{-31}) \times (1.38 \times 10^{-23}) \times 300 \]
Calculate this step by step:
- \(3 \times 9.1 \times 10^{-31} = 2.73 \times 10^{-30}\)
- \(2.73 \times 10^{-30} \times 1.38 \times 10^{-23} = 3.77 \times 10^{-53}\)
- \(3.77 \times 10^{-53} \times 300 = 1.131 \times 10^{-50}\)

Now take the square root:
\[ \sqrt{1.131 \times 10^{-50}} \approx 1.064 \times 10^{-25} \]

9. **Calculate the Wavelength**:
Now substitute this back into the wavelength formula:
\[ \lambda \approx \frac{6.63 \times 10^{-34}}{1.064 \times 10^{-25}} \approx 6.22 \times 10^{-9} \, m \]

10. **Final Result**:
Thus, the de Broglie wavelength of the electron in the metal at \(27^\circ C\) is approximately:
\[ \lambda \approx 6.22 \times 10^{-9} \, m \]