To produce characteristic `X` - rays using a Tungsten target in an X - ray generator , the accelerating should be greater than ……. Volts and the energy of the characterization is …… eV .
(The binding energy of the intermost electron in Tungsten is - 40keV).

Explanation

To solve the problem of determining the minimum accelerating voltage required to produce characteristic X-rays using a Tungsten target and the energy of the characteristic X-rays, we can follow these steps:

### Step 1: Understand the Binding Energy
The binding energy of the innermost electron in Tungsten is given as 40 keV. This value indicates the energy required to remove the electron from its orbit.

### Step 2: Determine the Energy of Characteristic X-rays
The energy of the emitted characteristic X-rays (E1) can be calculated using the relationship between the binding energy and the energy of the emitted X-ray. The energy of the characteristic X-ray is typically a fraction of the binding energy of the electron that was ejected.

Using the formula:
\[ E1 = \frac{3}{4} \times \text{Binding Energy} \]
Substituting the binding energy:
\[ E1 = \frac{3}{4} \times 40 \text{ keV} = 30 \text{ keV} \]
Thus, the energy of the characteristic X-ray is 30 keV.

### Step 3: Convert Energy to Electron Volts
Since 1 keV = 1000 eV, we convert 30 keV to eV:
\[ E1 = 30 \text{ keV} = 30,000 \text{ eV} \]

### Step 4: Determine the Minimum Accelerating Voltage
The minimum accelerating voltage (V) required to produce the characteristic X-rays can be found using the relationship:
\[ E = eV \]
Where:
- E is the energy of the X-ray in electron volts (eV)
- e is the charge of an electron (which cancels out in the equation)

Rearranging the formula gives:
\[ V = \frac{E1}{e} \]
Since we are using energy in eV, we can directly use:
\[ V = E1 \]
Thus:
\[ V = 30,000 \text{ volts} \]

### Final Answers
- The minimum accelerating voltage should be greater than **30,000 volts**.
- The energy of the characteristic X-ray is **30,000 eV**.

---