The `K_(alpha) X`-ray emission line of tungsten occurs at `lambda = 0.021 nm`. The energy difference between K and L levels in this aotm is about : `0.51 MeV`

Options

A. `0.51 MeV`
B. `1.2 MeV`
C. `59 keV`
D. `13.6 eV`

Explanation

To find the energy difference between the K and L levels in a tungsten atom corresponding to the K-alpha X-ray emission line at a wavelength of \( \lambda = 0.021 \, \text{nm} \), we can use the formula for energy associated with a photon:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 1: Convert the wavelength from nanometers to meters
Given \( \lambda = 0.021 \, \text{nm} \):

\[ \lambda = 0.021 \, \text{nm} = 0.021 \times 10^{-9} \, \text{m} = 2.1 \times 10^{-11} \, \text{m} \]

**Hint:** Remember to convert nanometers to meters by multiplying by \( 10^{-9} \).

### Step 2: Substitute values into the energy formula
Now we can substitute the values of \( h \), \( c \), and \( \lambda \) into the energy formula:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2.1 \times 10^{-11} \, \text{m}} \]

### Step 3: Calculate the energy
Calculating the numerator:

\[ hc = (6.626 \times 10^{-34})(3 \times 10^8) = 1.9878 \times 10^{-25} \, \text{Jm} \]

Now, substituting back into the energy equation:

\[ E = \frac{1.9878 \times 10^{-25}}{2.1 \times 10^{-11}} \approx 9.46 \times 10^{-15} \, \text{J} \]

### Step 4: Convert energy from joules to electron volts
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ E \, (\text{in eV}) = \frac{9.46 \times 10^{-15} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 59 \, \text{keV} \]

### Step 5: Convert keV to MeV
Since \( 1 \, \text{MeV} = 1000 \, \text{keV} \):

\[ E \, (\text{in MeV}) = \frac{59 \, \text{keV}}{1000} \approx 0.059 \, \text{MeV} \]

### Step 6: Conclusion
The energy difference between the K and L levels in tungsten is approximately \( 0.059 \, \text{MeV} \).

**Final Answer:** The energy difference between the K and L levels in tungsten is about \( 0.059 \, \text{MeV} \).

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