To find the cutoff wavelength in an X-ray tube experiment where the accelerating voltage is given as 1.24 MV, we can follow these steps:
### Step 1: Convert the Accelerating Voltage to Energy
The energy (E) of the electrons can be calculated from the accelerating voltage (V) using the formula:
\[ E = eV \]
Where:
- \( e \) is the charge of an electron, approximately \( 1.602 \times 10^{-19} \) coulombs.
- \( V \) is the accelerating voltage in volts.
Given:
\[ V = 1.24 \, \text{MV} = 1.24 \times 10^6 \, \text{V} \]
Now, substituting the values:
\[ E = (1.602 \times 10^{-19} \, \text{C}) \times (1.24 \times 10^6 \, \text{V}) \]
\[ E = 1.24 \times 10^6 \, \text{eV} \]
### Step 2: Use the Energy-Wavelength Relation
The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).
- \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \, \text{m/s} \).
- \( \lambda \) is the wavelength in meters.
### Step 3: Calculate \( hc \)
First, we calculate \( hc \):
\[ hc = (6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^8 \, \text{m/s}) \]
\[ hc \approx 1.9878 \times 10^{-25} \, \text{J m} \]
### Step 4: Convert Energy from eV to Joules
Since the energy we have is in electron volts, we need to convert it to joules.
Using:
\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]
Thus,
\[ E = 1.24 \times 10^6 \, \text{eV} = 1.24 \times 10^6 \times 1.602 \times 10^{-19} \, \text{J} \]
\[ E \approx 1.987 \times 10^{-13} \, \text{J} \]
### Step 5: Calculate the Cutoff Wavelength
Rearranging the energy-wavelength relation gives:
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{1.987 \times 10^{-13} \, \text{J}} \]
\[ \lambda \approx 1.0004 \times 10^{-12} \, \text{m} \]
### Step 6: Convert Wavelength to Nanometers
To convert meters to nanometers:
\[ 1 \, \text{m} = 10^9 \, \text{nm} \]
Thus,
\[ \lambda \approx 1.0004 \times 10^{-12} \, \text{m} = 1.0004 \times 10^{-12} \times 10^9 \, \text{nm} \]
\[ \lambda \approx 0.0010004 \, \text{nm} \]
\[ \lambda \approx 10^{-3} \, \text{nm} \]
### Final Answer
The cutoff wavelength is approximately \( 10^{-3} \, \text{nm} \).
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