What is the de Broglie wavelength of an electron accelerated from rest through a potential difference of V volts?

Options

A. `(12.3)/(sqrt(V))Å`
B. `(12.3)/(V)Å`
C. `(12.3)/(V^(2))Å`
D. None of these

Answer

To find the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of V volts, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
The momentum \(p\) of a particle can be expressed in terms of its kinetic energy (KE). The kinetic energy of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the particle and \(v\) is its velocity. The momentum \(p\) can also be expressed as:
\[ p = mv \]
From the kinetic energy formula, we can derive the velocity:
\[ v = \sqrt{\frac{2KE}{m}} \]
Thus, substituting this into the momentum formula gives:
\[ p = m \sqrt{\frac{2KE}{m}} = \sqrt{2m \cdot KE} \]

### Step 3: Calculate the kinetic energy of the electron
When an electron is accelerated through a potential difference \(V\), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = eV \]
where \(e\) is the charge of the electron (approximately \(1.6 \times 10^{-19}\) coulombs).

### Step 4: Substitute kinetic energy into the momentum formula
Substituting \(KE = eV\) into the momentum equation gives:
\[ p = \sqrt{2m \cdot eV} \]

### Step 5: Substitute momentum into the de Broglie wavelength formula
Now we can substitute this expression for momentum back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

### Step 6: Plug in the values
Using the known values:
- \(h = 6.626 \times 10^{-34} \, \text{Js}\) (Planck's constant)
- \(m = 9.11 \times 10^{-31} \, \text{kg}\) (mass of the electron)

We can express the wavelength as:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 9.11 \times 10^{-31} \cdot eV}} \]

### Step 7: Simplify the expression
This simplifies to:
\[ \lambda = \frac{12.3 \, \text{Å}}{\sqrt{V}} \]
where \(V\) is in volts, and \(12.3 \, \text{Å}\) is a constant derived from the calculations.

### Final Answer
Thus, the de Broglie wavelength of an electron accelerated from rest through a potential difference of \(V\) volts is:
\[ \lambda = \frac{12.3 \, \text{Å}}{\sqrt{V}} \]

---