In a TV tube the electron are accelerated by a potential difference of 10 kV. Then, their deBroglie wavelength is nearly-

Options

A. `1.2 Å`
B. `0.12 Å`
C. `12 Å`
D. `0.01 Å`

Answer

To find the de Broglie wavelength of electrons accelerated by a potential difference of 10 kV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given potential difference**:
The potential difference (V) is given as 10 kV. We need to convert this into volts:
\[ V = 10 \text{ kV} = 10 \times 1000 \text{ V} = 10000 \text{ V} \]

2. **Use 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 electron.

3. **Relate momentum to kinetic energy**:
The kinetic energy (KE) gained by the electron when accelerated through a potential difference \(V\) is given by:
\[ KE = eV \]
where \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \text{ C}\)). Thus, the kinetic energy becomes:
\[ KE = (1.6 \times 10^{-19} \text{ C}) \times (10000 \text{ V}) = 1.6 \times 10^{-15} \text{ J} \]

4. **Calculate the momentum**:
The kinetic energy is also related to momentum by the equation:
\[ KE = \frac{p^2}{2m} \]
Rearranging gives:
\[ p = \sqrt{2m \cdot KE} \]
The mass \(m\) of an electron is approximately \(9.11 \times 10^{-31} \text{ kg}\). Substituting the values:
\[ p = \sqrt{2 \times (9.11 \times 10^{-31} \text{ kg}) \times (1.6 \times 10^{-15} \text{ J})} \]

5. **Calculate \(p\)**:
\[ p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-15}} \approx \sqrt{2.917 \times 10^{-45}} \approx 5.4 \times 10^{-23} \text{ kg m/s} \]

6. **Substitute \(p\) back into the de Broglie wavelength formula**:
Using \(h = 6.63 \times 10^{-34} \text{ J s}\):
\[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.4 \times 10^{-23}} \approx 1.22 \times 10^{-11} \text{ m} \]

7. **Convert to Angstroms**:
Since \(1 \text{ Angstrom} = 10^{-10} \text{ m}\):
\[ \lambda \approx 1.22 \times 10^{-11} \text{ m} = 0.122 \text{ Angstroms} \]

### Conclusion:
The de Broglie wavelength of the electrons accelerated by a potential difference of 10 kV is approximately \(0.122 \text{ Angstroms}\).