If an X-ray tube operates at the voltage of 10kV, find the ratio of the de-broglie wavelength of the incident electrons to the shortest wavelength of X-ray producted. The specific charge of electron is `1.8xx10^11 C/kg`.

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced when the X-ray tube operates at a voltage of 10 kV.

### Step 1: Calculate the de Broglie wavelength of the incident electrons.

The formula for the de Broglie wavelength (λ_b) of a particle is given by:

\[ \lambda_b = \frac{h}{\sqrt{2qV \cdot m}} \]

Where:
- \( h \) is the Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( q \) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \( V \) is the voltage (10 kV = \(10 \times 10^3 \, \text{V}\)),
- \( m \) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)).

### Step 2: Substitute the values into the de Broglie wavelength formula.

First, we need to calculate \(2qV\):

\[ 2qV = 2 \times (1.6 \times 10^{-19} \, \text{C}) \times (10 \times 10^3 \, \text{V}) = 3.2 \times 10^{-15} \, \text{J} \]

Now, substituting into the de Broglie wavelength formula:

\[ \lambda_b = \frac{6.626 \times 10^{-34}}{\sqrt{3.2 \times 10^{-15} \cdot 9.11 \times 10^{-31}}} \]

### Step 3: Calculate the value under the square root.

Calculating \(3.2 \times 10^{-15} \cdot 9.11 \times 10^{-31}\):

\[ 3.2 \times 10^{-15} \cdot 9.11 \times 10^{-31} = 2.9152 \times 10^{-45} \]

Now, taking the square root:

\[ \sqrt{2.9152 \times 10^{-45}} \approx 5.396 \times 10^{-23} \]

### Step 4: Calculate the de Broglie wavelength.

Now substituting back into the wavelength formula:

\[ \lambda_b = \frac{6.626 \times 10^{-34}}{5.396 \times 10^{-23}} \approx 1.228 \times 10^{-11} \, \text{m} \]

### Step 5: Calculate the shortest wavelength of X-rays produced.

The formula for the shortest wavelength (λ_m) of X-rays produced is given by:

\[ \lambda_m = \frac{hc}{qV} \]

Where:
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)).

Substituting the values:

\[ \lambda_m = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{(1.6 \times 10^{-19})(10 \times 10^3)} \]

Calculating the denominator:

\[ (1.6 \times 10^{-19})(10 \times 10^3) = 1.6 \times 10^{-15} \]

Now substituting back into the wavelength formula:

\[ \lambda_m = \frac{1.9878 \times 10^{-25}}{1.6 \times 10^{-15}} \approx 1.241 \times 10^{-10} \, \text{m} \]

### Step 6: Calculate the ratio of the de Broglie wavelength to the shortest wavelength.

Now we can find the ratio:

\[ \text{Ratio} = \frac{\lambda_b}{\lambda_m} = \frac{1.228 \times 10^{-11}}{1.241 \times 10^{-10}} \approx 0.0989 \approx 0.1 \]

### Final Answer:

The ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced is approximately **0.1**.

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