To find the wavelength of an electron accelerated through a potential difference of 1 volt, we can use the de Broglie wavelength formula, which is given by:
\[
\lambda = \frac{h}{\sqrt{2 m_e e V}}
\]
Where:
- \(\lambda\) is the wavelength,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(m_e\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)),
- \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \(V\) is the potential difference (1 volt in this case).
### Step 1: Substitute the known values into the formula
Given:
- \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- \(m_e = 9.1 \times 10^{-31} \, \text{kg}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)
- \(V = 1 \, \text{V}\)
Substituting these values into the equation:
\[
\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 1}}
\]
### Step 2: Calculate the denominator
First, calculate the term inside the square root:
\[
2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} = 2.912 \times 10^{-49}
\]
Now, take the square root:
\[
\sqrt{2.912 \times 10^{-49}} = 5.39 \times 10^{-25}
\]
### Step 3: Calculate the wavelength
Now substitute this back into the wavelength formula:
\[
\lambda = \frac{6.63 \times 10^{-34}}{5.39 \times 10^{-25}} \approx 1.23 \times 10^{-9} \, \text{m}
\]
### Step 4: Convert to Angstroms
Since \(1 \, \text{Å} = 10^{-10} \, \text{m}\):
\[
\lambda \approx 12.3 \, \text{Å}
\]
### Final Answer
Thus, the wavelength of an electron accelerated through a potential difference of 1 volt is approximately \(12.3 \, \text{Å}\).
