To find the cutoff wavelength for a beam of 35.0 keV electrons striking a molybdenum target, we can follow these steps:
### Step 1: Understand the Concept of Cutoff Wavelength
The cutoff wavelength (λ_min) is the minimum wavelength of X-ray photons produced when electrons transfer all their energy to a single photon. This corresponds to the maximum energy of the photon generated.
### Step 2: Use the Energy-Wavelength Relationship
The energy of a photon is related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.0 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.
### Step 3: Convert the Energy from keV to Joules
The energy of the electrons is given as 35.0 keV. We need to convert this to Joules.
1 eV = \( 1.6 \times 10^{-19} \, \text{J} \), so:
\[ 35.0 \, \text{keV} = 35.0 \times 10^3 \, \text{eV} = 35.0 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J} \]
Calculating this gives:
\[ E = 5.6 \times 10^{-15} \, \text{J} \]
### Step 4: Rearrange the Energy-Wavelength Equation
To find the cutoff wavelength, rearrange the equation:
\[ \lambda = \frac{hc}{E} \]
### Step 5: Substitute the Values
Now substitute the values of \( h \), \( c \), and \( E \):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.0 \times 10^8 \, \text{m/s})}{5.6 \times 10^{-15} \, \text{J}} \]
### Step 6: Calculate the Wavelength
Perform the calculation:
\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{5.6 \times 10^{-15} \, \text{J}} \]
\[ \lambda \approx 3.55 \times 10^{-11} \, \text{m} \]
### Step 7: Convert to Picometers
Since \( 1 \, \text{pm} = 10^{-12} \, \text{m} \):
\[ \lambda \approx 35.5 \, \text{pm} \]
### Final Answer
The cutoff wavelength is approximately \( 35.5 \, \text{pm} \).
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