In an x-ray tube, electrons with energy 35 keV are incident on a cobalt `(Z=27)` target. Determine the cutoff wavelength for x-ray production.

Options

A. `1.4xx10^(-11)m`
B. `3.6xx10^(-11)m`
C. `1.8xx10^(-11)m`
D. `3.2xx10^(-11)m`

Answer

To determine the cutoff wavelength for X-ray production when electrons with energy 35 keV are incident on a cobalt target, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy of Electrons**:
The energy of the electrons is given as 35 keV. We need to convert this energy into joules for our calculations.
\[ E = 35 \, \text{keV} = 35 \times 10^3 \, \text{eV} \]
Since 1 eV = \(1.6 \times 10^{-19}\) joules, we can convert:
\[ E = 35 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J} \]

2. **Calculate the Energy in Joules**:
\[ E = 35 \times 1.6 \times 10^{-16} \, \text{J} = 5.6 \times 10^{-15} \, \text{J} \]

3. **Use the Energy-Wavelength Relationship**:
The relationship between energy (E) and wavelength (λ) is given by the equation:
\[ E = \frac{hc}{\lambda} \]
Rearranging this equation to solve for wavelength gives:
\[ \lambda = \frac{hc}{E} \]

4. **Substitute Constants**:
Here, \(h\) (Planck's constant) is \(6.63 \times 10^{-34} \, \text{J s}\) and \(c\) (speed of light) is \(3 \times 10^8 \, \text{m/s}\). Substituting these values into the equation:
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{5.6 \times 10^{-15} \, \text{J}} \]

5. **Calculate the Wavelength**:
\[ \lambda = \frac{1.989 \times 10^{-25} \, \text{J m}}{5.6 \times 10^{-15} \, \text{J}} \approx 3.55 \times 10^{-11} \, \text{m} \]

6. **Convert to Nanometers**:
To express the wavelength in more common units (nanometers), we convert:
\[ \lambda = 3.55 \times 10^{-11} \, \text{m} = 0.0355 \, \text{nm} \]

7. **Final Result**:
The cutoff wavelength for X-ray production is approximately \(3.55 \times 10^{-11} \, \text{m}\) or \(0.0355 \, \text{nm}\).