To solve the problem, we need to determine the potential \( V \) required to accelerate an electron beam so that its wavelength equals \( 1.0 \, \text{Å} \) (or \( 1.0 \times 10^{-10} \, \text{m} \)). We will use the relationship between kinetic energy, momentum, and wavelength.
### Step-by-Step Solution:
1. **Understanding the relationship between kinetic energy and potential:**
The kinetic energy (\( KE \)) gained by an electron when it is accelerated through a potential \( V \) is given by:
\[
KE = eV
\]
where \( e \) is the charge of the electron.
2. **Relating kinetic energy to momentum:**
The kinetic energy can also be expressed in terms of momentum (\( p \)):
\[
KE = \frac{p^2}{2m_e}
\]
where \( m_e \) is the mass of the electron.
3. **Using the de Broglie wavelength formula:**
The de Broglie wavelength (\( \lambda \)) of an electron is given by:
\[
\lambda = \frac{h}{p}
\]
where \( h \) is Planck's constant. Rearranging gives:
\[
p = \frac{h}{\lambda}
\]
4. **Substituting momentum into kinetic energy:**
Substitute \( p = \frac{h}{\lambda} \) into the kinetic energy expression:
\[
KE = \frac{1}{2m_e} \left(\frac{h}{\lambda}\right)^2
\]
5. **Setting the two expressions for kinetic energy equal:**
Since both expressions represent kinetic energy, we can set them equal:
\[
eV = \frac{1}{2m_e} \left(\frac{h}{\lambda}\right)^2
\]
6. **Rearranging to solve for \( V \):**
Rearranging gives:
\[
V = \frac{h^2}{2m_e e \lambda^2}
\]
7. **Substituting known values:**
We know:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m_e = 9.11 \times 10^{-31} \, \text{kg} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( \lambda = 1.0 \, \text{Å} = 1.0 \times 10^{-10} \, \text{m} \)
Plugging in these values:
\[
V = \frac{(6.626 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (1.0 \times 10^{-10})^2}
\]
8. **Calculating \( V \):**
First, calculate \( h^2 \):
\[
h^2 = (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{J}^2\text{s}^2
\]
Next, calculate the denominator:
\[
2 \times (9.11 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (1.0 \times 10^{-10})^2 = 2.91 \times 10^{-59} \, \text{J}^2
\]
Now, substituting these into the equation for \( V \):
\[
V = \frac{4.39 \times 10^{-67}}{2.91 \times 10^{-59}} \approx 1.51 \times 10^{2} \, \text{V} \approx 150 \, \text{V}
\]
### Final Answer:
Thus, the potential \( V \) required is approximately **150 volts**.
