A beam of electrons having energy 28 keV each strikes a target generating X rays. The minimum wave length `lambda_("min")` (called cut off wavelength ) of the X rays generated is

Options

A. 4.4 nm
B. 44 nm
C. 0.044 nm
D. 0.44 nm

Answer

To find the minimum wavelength (λ_min) of the X-rays generated when a beam of electrons strikes a target, we can use the following steps:

### Step-by-Step Solution:

1. **Understand the Energy of Electrons**:
The energy of the electrons is given as 28 keV (kiloelectron volts). We need to convert this energy into electron volts (eV) for our calculations:
\[ E = 28 \text{ keV} = 28,000 \text{ eV} \]

2. **Use the Energy-Wavelength Relation**:
The minimum wavelength of the X-rays can be calculated using the formula:
\[ \lambda_{\text{min}} = \frac{hc}{E} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \))
- \( E \) is the energy in joules.

3. **Convert Energy to Joules**:
To use the formula, we need to convert the energy from eV to joules. The conversion factor is:
\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]
Thus,
\[ E = 28,000 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 4.48 \times 10^{-15} \text{ J} \]

4. **Substitute Values into the Formula**:
Now we can substitute the values of \( h \), \( c \), and \( E \) into the formula for \( \lambda_{\text{min}} \):
\[ \lambda_{\text{min}} = \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{4.48 \times 10^{-15} \text{ J}} \]

5. **Calculate \( \lambda_{\text{min}} \)**:
Performing the calculation:
\[ \lambda_{\text{min}} = \frac{1.9878 \times 10^{-25} \text{ J m}}{4.48 \times 10^{-15} \text{ J}} \approx 4.44 \times 10^{-11} \text{ m} \]
Converting meters to nanometers (1 m = \( 10^9 \) nm):
\[ \lambda_{\text{min}} \approx 0.0444 \text{ nm} \]

6. **Final Result**:
Thus, the minimum wavelength (cut-off wavelength) of the X-rays generated is:
\[ \lambda_{\text{min}} \approx 0.044 \text{ nm} \]