An X ray tube is operated at an accelerating potential of 40 kV. What is the minimum wavelength of X rays produced1?

Options

A. 0.62 Å
B. 0.31 Å
C. 0.45 Å
D. 0.75 Å

Explanation

To find the minimum wavelength of X-rays produced by an X-ray tube operated at an accelerating potential of 40 kV, we can use the relationship between energy and wavelength. Here’s the step-by-step solution:

### Step 1: Understand the relationship between energy and wavelength
The energy (E) of a photon can be expressed in terms of its wavelength (λ) using the formula:
\[ E = \frac{hc}{\lambda} \]
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 (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 2: Convert the accelerating potential to energy
The energy gained by an electron when accelerated through a potential difference (V) is given by:
\[ E = eV \]
where:
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( V \) is the accelerating potential in volts.

Given \( V = 40 \, \text{kV} = 40,000 \, \text{V} \), we can calculate the energy:
\[ E = (1.6 \times 10^{-19} \, \text{C})(40,000 \, \text{V}) = 6.4 \times 10^{-15} \, \text{J} \]

### Step 3: Substitute the energy into the wavelength formula
Now, we can rearrange the energy-wavelength formula to find the minimum wavelength:
\[ \lambda = \frac{hc}{E} \]

### Step 4: Substitute the known values
Substituting the values of \( h \), \( c \), and \( E \):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{6.4 \times 10^{-15} \, \text{J}} \]

### Step 5: Calculate the wavelength
Calculating the above expression:
\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{6.4 \times 10^{-15} \, \text{J}} \approx 3.11 \times 10^{-11} \, \text{m} \]

### Step 6: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \( 1 \times 10^{-10} \, \text{m} \)):
\[ \lambda \approx 3.11 \times 10^{-11} \, \text{m} = 0.311 \, \text{nm} = 3.11 \, \text{Å} \]

### Final Answer
The minimum wavelength of X-rays produced is approximately \( 0.311 \, \text{Å} \).

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