The energy of a tungsten atom with a vacancy in L small is `11.3 KaV`. Wavelength of `K_(alpha)` photon for tungsten is `21.3 "pm"`. If a potential difference of `62 kV` is applied across the `X`-raystube following characterstic X-rays will be produced.

Options

A. K,L series
B. only `K_(alpha)` & `L` series
C. only L series
D. None of these

Explanation

To solve the problem, we need to analyze the energy levels and the characteristics of X-rays produced by tungsten when a potential difference is applied. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Energy of tungsten atom with a vacancy in L shell: \( E_1 = 11.3 \, \text{keV} \)
- Wavelength of \( K_{\alpha} \) photon: \( \lambda = 21.3 \, \text{pm} = 21.3 \times 10^{-12} \, \text{m} \)
- Potential difference applied: \( V = 62 \, \text{kV} \)

### Step 2: Calculate the energy of the \( K_{\alpha} \) photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

First, we need to convert the wavelength from picometers to meters:
\[ \lambda = 21.3 \, \text{pm} = 21.3 \times 10^{-12} \, \text{m} \]

Now, substituting the values into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{21.3 \times 10^{-12} \, \text{m}} \]

Calculating this gives:
\[ E \approx \frac{1.9878 \times 10^{-25}}{21.3 \times 10^{-12}} \approx 9.33 \times 10^{-15} \, \text{J} \]

### Step 3: Convert the energy from Joules to keV
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \approx \frac{9.33 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 58.21 \, \text{keV} \]

### Step 4: Calculate the change in energy
The change in energy when an electron transitions from the L shell to the K shell can be calculated as:
\[ \Delta E = E_2 - E_1 \]
Where:
- \( E_2 = 58.21 \, \text{keV} \)
- \( E_1 = 11.3 \, \text{keV} \)

Thus:
\[ \Delta E = 58.21 \, \text{keV} - 11.3 \, \text{keV} = 46.91 \, \text{keV} \]

### Step 5: Compare with the applied potential difference
The maximum energy of the X-rays produced is equal to the energy corresponding to the applied potential difference:
\[ E_{\text{max}} = eV = 62 \, \text{keV} \]

Since \( \Delta E = 46.91 \, \text{keV} < 62 \, \text{keV} \), we conclude that the transition can occur.

### Step 6: Determine the series of X-rays produced
Since the change in energy is less than the applied potential difference, only the L series will be produced.

### Conclusion
The correct option is **C**, which indicates that only the L series of X-rays will be produced.

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