Four the production of X-rays of wavelength 0.1 Å the minimum potential difference will be

Options

A. 12.4 kV
B. 24.8 kV
C. 124 kV
D. 248 kV

Answer

To find the minimum potential difference required for the production of X-rays with a wavelength of 0.1 Å, we can use the relationship between the energy of the X-rays and the potential difference applied to accelerate the electrons. Here’s a step-by-step solution:

### Step 1: Understand the relationship between energy, wavelength, and potential difference
The energy (E) of a photon (in this case, X-ray) is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda \) is the wavelength (in meters)

### Step 2: Convert the wavelength from Ångstroms to meters
Given that the wavelength \( \lambda = 0.1 \, \text{Å} \):
\[ \lambda = 0.1 \, \text{Å} = 0.1 \times 10^{-10} \, \text{m} = 1.0 \times 10^{-11} \, \text{m} \]

### Step 3: Calculate the energy of the X-ray photon
Substituting the values into the energy equation:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{1.0 \times 10^{-11} \, \text{m}} \]

### Step 4: Perform the calculation
Calculating the numerator:
\[ E = \frac{(6.626 \times 3) \times 10^{-34 + 8}}{1.0 \times 10^{-11}} \]
\[ E = \frac{19.878 \times 10^{-26}}{1.0 \times 10^{-11}} \]
\[ E = 19.878 \times 10^{-15} \, \text{J} \]

### Step 5: Relate the energy to the potential difference
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 (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 6: Solve for the potential difference \( V \)
Setting the two expressions for energy equal:
\[ eV = 19.878 \times 10^{-15} \]
\[ V = \frac{19.878 \times 10^{-15}}{1.6 \times 10^{-19}} \]

### Step 7: Calculate the potential difference
\[ V = \frac{19.878}{1.6} \times 10^{4} \]
\[ V \approx 12.3675 \times 10^{4} \]
\[ V \approx 123.67 \, \text{kV} \]

### Conclusion
The minimum potential difference required for the production of X-rays of wavelength 0.1 Å is approximately **124 kV**.

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