An electron is accelerated through a potential difference of `10,000V`. Its de Broglie wavelength is, (nearly): `(me=9xx10^(-31)kg)`

Options

A. 12.2nm
B. `12.2xx10^(-13)m`
C. `12.2xx10^(-12)m`
D. `12.2xx10^(-14)m`

Explanation

To find the de Broglie wavelength of an electron accelerated through a potential difference of \(10,000V\), we can follow these steps:

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

### Step 2: Calculate the momentum 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 (\(e \approx 1.6 \times 10^{-19} C\)) and \(V\) is the potential difference.

The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
where \(m\) is the mass of the electron. Setting these two expressions for kinetic energy equal gives:
\[ eV = \frac{p^2}{2m} \]

Rearranging this to find momentum \(p\):
\[ p = \sqrt{2m eV} \]

### Step 3: Substitute known values
Given:
- \(m_e = 9 \times 10^{-31} \, \text{kg}\)
- \(V = 10,000 \, \text{V}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)

Substituting these values into the momentum equation:
\[ p = \sqrt{2 \times (9 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (10,000)} \]

### Step 4: Calculate the momentum
Calculating the values:
\[ p = \sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10,000} \]
\[ = \sqrt{2 \times 9 \times 1.6 \times 10^{-31} \times 10^{-15}} \]
\[ = \sqrt{28.8 \times 10^{-15}} \approx 5.37 \times 10^{-8} \, \text{kg m/s} \]

### Step 5: Calculate the de Broglie wavelength
Now, using the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \(h \approx 6.63 \times 10^{-34} \, \text{Js}\).

Substituting \(p\):
\[ \lambda = \frac{6.63 \times 10^{-34}}{5.37 \times 10^{-8}} \]
Calculating this gives:
\[ \lambda \approx 1.23 \times 10^{-26} \, \text{m} \]

### Conclusion
Thus, the de Broglie wavelength of the electron is approximately \(1.23 \times 10^{-26} \, \text{m}\).