In the Coolidge tube experiment, if the applied voltage is increased to three times, the short wavelength limit of continuous X- ray spectrum shift by 20 pm. What is the initial voltage (in kV) applied to the tube?

Answer

To solve the problem, we need to determine the initial voltage (V) applied to the Coolidge tube, given that increasing the voltage to three times results in a decrease in the short wavelength limit of the continuous X-ray spectrum by 20 picometers.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The minimum wavelength (λ_min) of the X-ray spectrum is related to the voltage (V) applied to the Coolidge tube. The relationship is given by:
\[ \lambda_{\text{min}} = \frac{hc}{eV} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( c \) 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 voltage in volts.

2. **Initial Wavelength**:
Let the initial voltage be \( V \). The initial minimum wavelength is:
\[ \lambda_{\text{min}} = \frac{hc}{eV} \]

3. **Wavelength After Voltage Increase**:
When the voltage is increased to three times the initial voltage, the new minimum wavelength (λ_min') becomes:
\[ \lambda_{\text{min'}} = \frac{hc}{e(3V)} = \frac{hc}{3eV} \]

4. **Change in Wavelength**:
According to the problem, the change in wavelength is given as:
\[ \lambda_{\text{min}} - \lambda_{\text{min'}} = 20 \, \text{pm} = 20 \times 10^{-12} \, \text{m} \]
Substituting the expressions for λ_min and λ_min':
\[ \frac{hc}{eV} - \frac{hc}{3eV} = 20 \times 10^{-12} \]

5. **Simplifying the Equation**:
Factor out \( \frac{hc}{eV} \):
\[ \frac{hc}{eV} \left(1 - \frac{1}{3}\right) = 20 \times 10^{-12} \]
This simplifies to:
\[ \frac{hc}{eV} \cdot \frac{2}{3} = 20 \times 10^{-12} \]

6. **Solving for V**:
Rearranging gives:
\[ \frac{hc}{eV} = 30 \times 10^{-12} \]
Thus,
\[ V = \frac{hc}{30 \times 10^{-12} \cdot e} \]

7. **Substituting Values**:
Substitute \( h \), \( c \), and \( e \):
\[ V = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{30 \times 10^{-12} \cdot (1.6 \times 10^{-19})} \]

8. **Calculating V**:
Calculate the numerator:
\[ 6.63 \times 10^{-34} \times 3 \times 10^8 = 1.989 \times 10^{-25} \]
And the denominator:
\[ 30 \times 10^{-12} \times 1.6 \times 10^{-19} = 4.8 \times 10^{-30} \]
Therefore,
\[ V = \frac{1.989 \times 10^{-25}}{4.8 \times 10^{-30}} \approx 4.14 \times 10^4 \, \text{V} \]

9. **Converting to kV**:
Convert volts to kilovolts:
\[ V \approx 41.4 \, \text{kV} \]

### Final Answer:
The initial voltage applied to the tube is approximately **41.4 kV**.