A potential of 10000 V is applied across an x-ray tube. Find the ratio of de-Broglie wavelength associated with incident electrons to the minimum wavelength associated with x-rays.

Options

A. `10`
B. `20`
C. `1//10`
D. `1//20`

Explanation

To solve the problem of finding the ratio of the de-Broglie wavelength associated with incident electrons to the minimum wavelength associated with x-rays, we can follow these steps:

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

The de-Broglie wavelength (\( \lambda_E \)) of an electron can be calculated using the formula:

\[ \lambda_E = \frac{h}{p} \]

where \( h \) is Planck's constant and \( p \) is the momentum of the electron. The momentum can also be expressed in terms of kinetic energy (\( KE \)):

\[ p = \sqrt{2m_e KE} \]

Given that the potential difference \( V \) across the x-ray tube is 10,000 V, the kinetic energy of the electron is:

\[ KE = eV = 10,000 \, \text{eV} \]

Converting this to joules (using \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)):

\[ KE = 10,000 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-15} \, \text{J} \]

Substituting this into the momentum formula:

\[ p = \sqrt{2m_e \cdot KE} \]

Using \( m_e = 9.11 \times 10^{-31} \, \text{kg} \):

\[ p = \sqrt{2 \cdot (9.11 \times 10^{-31} \, \text{kg}) \cdot (1.6 \times 10^{-15} \, \text{J})} \]

Calculating this gives:

\[ p \approx \sqrt{2.91 \times 10^{-45}} \approx 5.39 \times 10^{-23} \, \text{kg m/s} \]

Now substituting back to find \( \lambda_E \):

\[ \lambda_E = \frac{h}{p} = \frac{6.63 \times 10^{-34} \, \text{Js}}{5.39 \times 10^{-23} \, \text{kg m/s}} \approx 1.23 \times 10^{-11} \, \text{m} = 0.122 \, \text{nm} \]

### Step 2: Calculate the minimum wavelength of the x-rays

The minimum wavelength (\( \lambda_P \)) of x-rays can be found using the energy-wavelength relationship:

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

For x-rays, the maximum energy corresponds to the potential difference applied:

\[ E = eV = 10,000 \, \text{eV} \]

Thus,

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

Converting \( E \) to joules:

\[ E = 10,000 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-15} \, \text{J} \]

Substituting \( h = 6.63 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):

\[ \lambda_P = \frac{(6.63 \times 10^{-34} \, \text{Js}) \cdot (3 \times 10^8 \, \text{m/s})}{1.6 \times 10^{-15} \, \text{J}} \approx \frac{1.989 \times 10^{-25}}{1.6 \times 10^{-15}} \approx 1.24 \times 10^{-10} \, \text{m} = 0.0124 \, \text{nm} \]

### Step 3: Calculate the ratio of the wavelengths

Now, we can find the ratio of the de-Broglie wavelength of the incident electrons to the minimum wavelength of the x-rays:

\[ \text{Ratio} = \frac{\lambda_E}{\lambda_P} = \frac{0.122 \, \text{nm}}{0.0124 \, \text{nm}} \approx 9.84 \approx 10 \]

### Final Answer

The ratio of the de-Broglie wavelength associated with incident electrons to the minimum wavelength associated with x-rays is approximately **10**.

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