The de Broglie wavelength of an electron is `1A^(@)`. In what potential difference electron accelerates

Options

A. 100 V
B. 150
C. 225 V
D. 256 V

Answer

To find the potential difference through which an electron accelerates to achieve a de Broglie wavelength of \(1 \, \text{Å}\), we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\(\lambda\)) 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 potential difference
For an electron accelerated through a potential difference \(V\), the kinetic energy gained is equal to the work done by the electric field:
\[ KE = eV \]
where \(e\) is the charge of the electron. The momentum \(p\) of the electron can also be expressed in terms of its mass \(m\) and velocity \(v\):
\[ p = mv \]
Using the relation between kinetic energy and momentum, we have:
\[ KE = \frac{p^2}{2m} \]

### Step 3: Substitute for momentum in terms of potential difference
From the kinetic energy equation:
\[ eV = \frac{p^2}{2m} \]
Rearranging gives:
\[ p = \sqrt{2m eV} \]

### Step 4: Substitute momentum into the de Broglie wavelength equation
Substituting \(p\) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m eV}} \]

### Step 5: Solve for potential difference \(V\)
Rearranging the equation to solve for \(V\):
\[ V = \frac{h^2}{2m e \lambda^2} \]

### Step 6: Plug in the values
1. **Planck's constant \(h\)**: \(6.626 \times 10^{-34} \, \text{Js}\)
2. **Mass of electron \(m\)**: \(9.11 \times 10^{-31} \, \text{kg}\)
3. **Charge of electron \(e\)**: \(1.602 \times 10^{-19} \, \text{C}\)
4. **Wavelength \(\lambda\)**: \(1 \, \text{Å} = 1 \times 10^{-10} \, \text{m}\)

Now substituting these values into the equation:
\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (1.602 \times 10^{-19}) \times (1 \times 10^{-10})^2} \]

### Step 7: Calculate the value
Calculating the numerator:
\[ (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \]
Calculating the denominator:
\[ 2 \times (9.11 \times 10^{-31}) \times (1.602 \times 10^{-19}) \times (1 \times 10^{-20}) = 2.91 \times 10^{-69} \]
Now substituting:
\[ V = \frac{4.39 \times 10^{-67}}{2.91 \times 10^{-69}} \approx 150.5 \, \text{V} \]

### Conclusion
The potential difference \(V\) through which the electron accelerates is approximately \(150 \, \text{V}\).