If the short wavelength limit of the continous spectrum coming out of a Coolidge tube is `10 A`, then the de Broglie wavelength of the electrons reaching the target netal in the Coolidge tube is approximately

Options

A. `0.3A`
B. `3A`
C. `30A`
D. `10A`

Explanation

To solve the problem, we need to find the de Broglie wavelength of the electrons reaching the target metal in a Coolidge tube, given that the short wavelength limit of the continuous spectrum is 10 angstroms (10 Å).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The energy of the photons corresponding to the short wavelength limit can be expressed using the formula:
\[ E = \frac{hc}{\lambda_{\text{min}}} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda_{\text{min}}\) is the minimum wavelength (10 Å or \(10 \times 10^{-10} \, \text{m}\)).

2. **Calculating the Energy**:
Substitute the values into the energy equation:
\[ E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{10 \times 10^{-10}} \]
Simplifying this gives:
\[ E = \frac{1.989 \times 10^{-25}}{10 \times 10^{-10}} = 1.989 \times 10^{-15} \, \text{J} \]

3. **Relating Energy to Kinetic Energy**:
The energy of the electrons is equal to their kinetic energy, which can be expressed as:
\[ E = \frac{p^2}{2m} \]
where \(p\) is the momentum and \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)).

4. **Finding Momentum**:
Rearranging the kinetic energy equation gives:
\[ p = \sqrt{2mE} \]
Substituting the values:
\[ p = \sqrt{2 \times (9.11 \times 10^{-31}) \times (1.989 \times 10^{-15})} \]
Calculating this yields:
\[ p \approx \sqrt{3.619 \times 10^{-45}} \approx 1.90 \times 10^{-22} \, \text{kg m/s} \]

5. **Using de Broglie Wavelength Formula**:
The de Broglie wavelength \(\lambda\) is given by:
\[ \lambda = \frac{h}{p} \]
Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{1.90 \times 10^{-22}} \]
Simplifying this gives:
\[ \lambda \approx 3.49 \times 10^{-12} \, \text{m} = 0.349 \, \text{nm} = 3.49 \, \text{Å} \]

6. **Final Answer**:
The de Broglie wavelength of the electrons is approximately \(0.34 \, \text{Å}\).

### Conclusion:
The correct option is the one that matches \(0.34 \, \text{Å}\).