To find the approximate kinetic energy of an electron such that its de-Broglie wavelength is equal to the wavelength of X-ray of maximum energy produced in an X-ray tube operating at 24800 V, we can follow these steps:
### Step 1: Calculate the wavelength of the X-ray
The maximum energy (E) of the X-ray produced in the tube can be calculated using the formula:
\[ E = eV \]
where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C),
- \( V \) is the potential difference (24800 V).
So,
\[ E = 1.6 \times 10^{-19} \, \text{C} \times 24800 \, \text{V} \]
\[ E = 3.968 \times 10^{-15} \, \text{J} \]
### Step 2: Relate energy to wavelength
Using the relationship between energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
we can rearrange this to find the wavelength (\( \lambda \)):
\[ \lambda = \frac{hc}{E} \]
Substituting \( h = 6.63 \times 10^{-34} \, \text{J s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \lambda = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{3.968 \times 10^{-15}} \]
Calculating this gives:
\[ \lambda \approx 5.00 \times 10^{-11} \, \text{m} \]
### Step 3: Set up the de-Broglie wavelength equation
The de-Broglie wavelength (\( \lambda \)) of an electron is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can also be expressed in terms of kinetic energy (K):
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)).
### Step 4: Equate the two wavelengths
Setting the two expressions for wavelength equal gives:
\[ \frac{h}{\sqrt{2mK}} = \lambda \]
Rearranging for kinetic energy (K):
\[ K = \frac{h^2}{2m\lambda^2} \]
### Step 5: Substitute known values
Substituting \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( m = 9.1 \times 10^{-31} \, \text{kg} \), and \( \lambda = 5.00 \times 10^{-11} \, \text{m} \):
\[ K = \frac{(6.63 \times 10^{-34})^2}{2 \times (9.1 \times 10^{-31}) \times (5.00 \times 10^{-11})^2} \]
Calculating this gives:
\[ K \approx 3.79 \times 10^{-14} \, \text{J} \]
### Step 6: Convert kinetic energy to electron volts
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ K \approx \frac{3.79 \times 10^{-14}}{1.6 \times 10^{-19}} \approx 236.25 \, \text{eV} \]
### Final Answer
The approximate kinetic energy of the electron should be around **236.25 eV**.
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