To find the wavelength of X-rays emitted when an electron jumps from the L-shell to the K-shell in tungsten, we can follow these steps:
### Step 1: Determine the energy levels of the K and L electrons
The energy of the K-electron in tungsten is given as:
\[ E_K = -20.0 \, \text{keV} \]
The energy of the L-electron is given as:
\[ E_L = -2.0 \, \text{keV} \]
### Step 2: Calculate the energy difference (ΔE)
The energy difference when an electron jumps from the L-shell to the K-shell can be calculated using the formula:
\[ \Delta E = E_{final} - E_{initial} \]
Here, the final state is the K-shell and the initial state is the L-shell:
\[ \Delta E = E_K - E_L \]
Substituting the values:
\[ \Delta E = (-20.0 \, \text{keV}) - (-2.0 \, \text{keV}) \]
\[ \Delta E = -20.0 + 2.0 \]
\[ \Delta E = -18.0 \, \text{keV} \]
### Step 3: Convert energy difference to joules
To use the energy in the formula for wavelength, we need to convert keV to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, for keV:
\[ 1 \, \text{keV} = 1.6 \times 10^{-16} \, \text{J} \]
Now converting ΔE:
\[ \Delta E = 18.0 \, \text{keV} \times 1.6 \times 10^{-16} \, \text{J/keV} \]
\[ \Delta E = 28.8 \times 10^{-16} \, \text{J} \]
\[ \Delta E = 2.88 \times 10^{-15} \, \text{J} \]
### Step 4: Use the energy-wavelength relationship
The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) is the energy in joules,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.
Rearranging the formula to solve for wavelength (\( \lambda \)):
\[ \lambda = \frac{hc}{E} \]
### Step 5: Substitute the values into the equation
Substituting the values of \( h \), \( c \), and \( \Delta E \):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{2.88 \times 10^{-15} \, \text{J}} \]
### Step 6: Calculate the wavelength
Calculating the numerator:
\[ hc = 6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25} \, \text{J m} \]
Now substituting into the wavelength equation:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{2.88 \times 10^{-15}} \]
\[ \lambda \approx 6.90 \times 10^{-11} \, \text{m} \]
### Step 7: Convert to angstroms
To convert meters to angstroms (1 Å = \( 10^{-10} \) m):
\[ \lambda \approx 0.690 \, \text{Å} \]
### Final Answer
The wavelength of the X-rays emitted when an electron jumps from the L-shell to the K-shell in tungsten is approximately:
\[ \lambda \approx 0.690 \, \text{Å} \]
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