x-rays are produced in an X-ray tube by electrons accelerated through an electric potential difference fo 50.0 kV. An electron makes three collisions in the target coming to rest and loses half its remaining kinetic energy in each of the first two collisions. Determine the wavelength of the resulting photons. (Neglecting the recoil of the heavy target atoms).

Explanation

To solve the problem of determining the wavelength of the resulting photons produced by an electron in an X-ray tube, we will follow these steps:

### Step 1: Calculate the initial kinetic energy of the electron
The kinetic energy (KE) of the electron accelerated through a potential difference (V) of 50 kV can be calculated using the formula:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C) and \( V \) is the potential difference in volts.

Substituting the values:
\[ KE = (1.6 \times 10^{-19} \, \text{C}) \cdot (50 \times 10^3 \, \text{V}) = 8.0 \times 10^{-15} \, \text{J} \]
Alternatively, in electron volts (eV), the kinetic energy is:
\[ KE = 50 \, \text{keV} = 50 \times 10^3 \, \text{eV} \]

### Step 2: Determine the kinetic energy after the first collision
After the first collision, the electron loses half of its kinetic energy:
\[ KE_1 = \frac{1}{2} KE = \frac{1}{2} \times 50 \, \text{keV} = 25 \, \text{keV} \]

### Step 3: Determine the kinetic energy after the second collision
After the second collision, the electron again loses half of its remaining kinetic energy:
\[ KE_2 = \frac{1}{2} KE_1 = \frac{1}{2} \times 25 \, \text{keV} = 12.5 \, \text{keV} \]

### Step 4: Determine the kinetic energy after the third collision
After the third collision, the electron comes to rest, so:
\[ KE_3 = 0 \, \text{keV} \]

### Step 5: Calculate the wavelength of the resulting photons after each collision
The wavelength of the emitted X-ray photon can be calculated using the formula:
\[ \lambda = \frac{12375}{E} \]
where \( E \) is the energy in eV.

#### Wavelength after the first collision:
\[ \lambda_1 = \frac{12375}{25} = 495 \, \text{Å} \, (\text{or } 0.495 \, \text{nm}) \]

#### Wavelength after the second collision:
\[ \lambda_2 = \frac{12375}{12.5} = 990 \, \text{Å} \, (\text{or } 0.99 \, \text{nm}) \]

### Final Result
The wavelengths of the resulting photons after the first and second collisions are:
- After the first collision: \( 495 \, \text{Å} \)
- After the second collision: \( 990 \, \text{Å} \)