in order that the short wavelength limit of the continuous X-ray spectrum be `1 Å`, the potential difference through which an electron must be accelerated is

Options

A. 124 kV
B. 1.24 kV
C. 12.4 kV
D. 1240 kV

Explanation

To solve the problem of finding the potential difference required to achieve a short wavelength limit of 1 Å in the continuous X-ray spectrum, we can follow these steps:

### Step 1: Understand the relationship between wavelength and potential difference
The minimum wavelength (λ_min) of the continuous X-ray spectrum is related to the accelerating potential (V) of the electrons by the formula:
\[ \lambda_{\text{min}} = \frac{hc}{eV} \]
where:
- \( \lambda_{\text{min}} \) is the minimum wavelength,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( e \) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \( V \) is the potential difference in volts.

### Step 2: Convert the wavelength from Ångstroms to meters
Given that \( \lambda_{\text{min}} = 1 \, \text{Å} \):
\[ 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \]

### Step 3: Rearrange the formula to solve for V
Rearranging the formula gives:
\[ V = \frac{hc}{e \lambda_{\text{min}}} \]

### Step 4: Substitute the known values into the equation
Substituting the constants and the value of \( \lambda_{\text{min}} \):
\[ V = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (1 \times 10^{-10} \, \text{m})} \]

### Step 5: Calculate the numerator and denominator
Calculating the numerator:
\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Jm} \]

Calculating the denominator:
\[ 1.6 \times 10^{-19} \times 1 \times 10^{-10} = 1.6 \times 10^{-29} \, \text{C m} \]

### Step 6: Divide the numerator by the denominator
Now, we can calculate \( V \):
\[ V = \frac{1.9878 \times 10^{-25}}{1.6 \times 10^{-29}} = 124.24 \times 10^3 \, \text{V} = 124.24 \, \text{kV} \]

### Step 7: Round the answer to the appropriate significant figures
Rounding \( 124.24 \, \text{kV} \) gives approximately \( 124 \, \text{kV} \).

### Conclusion
Thus, the potential difference through which an electron must be accelerated to achieve a short wavelength limit of 1 Å is approximately:
\[ \boxed{124 \, \text{kV}} \]