An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength is equal to 1.6 A?

Options

A. `58.90 V`
B. `85.75V`
C. `45.35`V
D. `105.31` V

Answer

To solve the problem of determining the potential through which a beam of electrons should be accelerated to achieve a wavelength of 1.6 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between energy and potential:**
The kinetic energy (KE) of the electrons after being accelerated through a potential \( V \) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs).

2. **Use the de Broglie wavelength formula:**
The de Broglie wavelength \( \lambda \) of a particle is given by:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant (approximately \( 6.63 \times 10^{-34} \) J·s), \( m \) is the mass of the electron (approximately \( 9.1 \times 10^{-31} \) kg), and \( v \) is the velocity of the electron.

3. **Relate kinetic energy to velocity:**
The kinetic energy can also be expressed in terms of velocity:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ eV = \frac{1}{2} mv^2 \]

4. **Express velocity in terms of wavelength:**
From the de Broglie equation, we can express \( v \) as:
\[ v = \frac{h}{m\lambda} \]
Substitute this expression for \( v \) into the kinetic energy equation:
\[ eV = \frac{1}{2} m \left(\frac{h}{m\lambda}\right)^2 \]

5. **Simplify the equation:**
Rearranging gives:
\[ eV = \frac{1}{2} \frac{h^2}{m\lambda^2} \]
Therefore, we can express \( V \) as:
\[ V = \frac{h^2}{2em\lambda^2} \]

6. **Substitute the known values:**
Now plug in the values:
- \( h = 6.63 \times 10^{-34} \) J·s
- \( e = 1.6 \times 10^{-19} \) C
- \( m = 9.1 \times 10^{-31} \) kg
- \( \lambda = 1.6 \) Å = \( 1.6 \times 10^{-10} \) m

Substituting these into the equation:
\[ V = \frac{(6.63 \times 10^{-34})^2}{2 \times (1.6 \times 10^{-19}) \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-10})^2} \]

7. **Calculate the potential:**
Performing the calculations step-by-step:
- Calculate \( h^2 \):
\[ h^2 = (6.63 \times 10^{-34})^2 \approx 4.39 \times 10^{-67} \text{ J}^2\text{s}^2 \]
- Calculate \( (1.6 \times 10^{-10})^2 \):
\[ (1.6 \times 10^{-10})^2 = 2.56 \times 10^{-20} \text{ m}^2 \]
- Substitute and simplify:
\[ V \approx \frac{4.39 \times 10^{-67}}{2 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31} \times 2.56 \times 10^{-20}} \]
- After calculating the denominator and performing the division, we find:
\[ V \approx 58.9 \text{ volts} \]

### Final Answer:
The potential through which the beam of electrons should be accelerated is approximately **58.9 volts**.