To solve the problem of finding the minimum wavelength of X-rays generated when a potential difference of 20 kV is applied across an X-ray tube, we can use the relationship between energy and wavelength. Here's the step-by-step solution:
### Step 1: Understand the relationship between energy and wavelength
The energy (E) of the X-rays produced can be expressed in terms of the charge (e) and the potential difference (V):
\[ E = eV \]
where:
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \) coulombs)
- \( V \) is the potential difference in volts
### Step 2: Convert the potential difference
The potential difference given is 20 kV. We convert this to volts:
\[ V = 20 \, \text{kV} = 20 \times 10^3 \, \text{V} = 2 \times 10^4 \, \text{V} \]
### Step 3: Calculate the energy in joules
Using the charge of an electron, we can calculate the energy:
\[ E = eV = (1.6 \times 10^{-19} \, \text{C})(2 \times 10^4 \, \text{V}) \]
\[ E = 3.2 \times 10^{-15} \, \text{J} \]
### Step 4: Relate energy to wavelength
The energy of a photon can also be expressed in terms of its wavelength (\( \lambda \)):
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{J s} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
### Step 5: Rearrange the equation to find wavelength
Rearranging the equation for wavelength gives:
\[ \lambda = \frac{hc}{E} \]
### Step 6: Substitute known values
Substituting the values of \( h \), \( c \), and \( E \):
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{3.2 \times 10^{-15} \, \text{J}} \]
### Step 7: Calculate the wavelength
Calculating the above expression:
\[ \lambda = \frac{1.989 \times 10^{-25}}{3.2 \times 10^{-15}} \]
\[ \lambda \approx 6.22 \times 10^{-11} \, \text{m} \]
### Step 8: Convert to angstroms
Since \( 1 \, \text{angstrom} = 10^{-10} \, \text{m} \):
\[ \lambda \approx 0.622 \, \text{angstroms} \]
### Final Answer
The minimum wavelength of X-rays generated is approximately:
\[ \lambda \approx 0.62 \, \text{angstroms} \]
