The `K_(alpha)` X-ray emission line of lungsten accurs at `lambda = 0.021 nm`. What is the energy difference between `K`and `L` levels in the atom?

Options

A. `0.51 meV`
B. `1.2 meV`
C. `59 meV`
D. `13.6 meV`

Explanation

To find the energy difference between the K and L levels in the tungsten atom based on the K_alpha X-ray emission line, we can follow these steps:

### Step 1: Understand the K_alpha X-ray emission
The K_alpha X-ray is produced when an electron transitions from the L level (n=2) to the K level (n=1). The energy of the emitted X-ray corresponds to the energy difference between these two levels.

### Step 2: Use the formula for energy
The energy (E) of the emitted X-ray can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) 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 of the X-ray in meters.

### Step 3: Convert the wavelength to meters
Given the wavelength \( \lambda = 0.021 \, \text{nm} \), we convert it to meters:
\[ \lambda = 0.021 \, \text{nm} = 0.021 \times 10^{-9} \, \text{m} \]

### Step 4: Substitute the values into the energy formula
Now, substituting the values into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^{8} \, \text{m/s})}{0.021 \times 10^{-9} \, \text{m}} \]

### Step 5: Calculate the energy in Joules
Calculating this gives:
\[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{0.021 \times 10^{-9}} \]
\[ E \approx 9.46 \times 10^{-14} \, \text{J} \]

### Step 6: Convert energy from Joules to electron volts
To convert the energy from Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \text{ (in eV)} = \frac{9.46 \times 10^{-14} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]
\[ E \approx 591.25 \, \text{eV} \]

### Step 7: Final answer
The energy difference between the K and L levels in the tungsten atom is approximately:
\[ E \approx 591.25 \, \text{eV} \]