The voltage applied to the Coolidge X-ray tube is increased by 25%. As a result the short wave limit of continuous X-ray spectrum shifts by `Delta lambda`. The initial voltage applied to the tube is

Options

A. `(hc)/(4e Delta lambda)`
B. `(hc)/(5e Delta lambda)`
C. `(4hc)/(5e Delta lambda)`
D. `(5hc)/(4e Delta lambda)`

Answer

To solve the problem, we need to analyze the relationship between the voltage applied to the Coolidge X-ray tube and the wavelength of the emitted X-rays. The key concept here is that the wavelength of the emitted X-rays is inversely proportional to the voltage applied.

### Step-by-Step Solution:

1. **Understanding the relationship**:
The energy of the emitted X-rays is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where \(E\) is the energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.

2. **Relating energy to voltage**:
The energy of the X-rays can also be expressed in terms of the voltage \(V\) applied to the tube:
\[ E = eV \]
where \(e\) is the charge of an electron.

3. **Setting up the equations**:
For the initial voltage \(V_1\), we have:
\[ \lambda_1 = \frac{hc}{eV_1} \]
For the increased voltage \(V_2\), which is 25% more than \(V_1\):
\[ V_2 = V_1 + 0.25V_1 = \frac{5}{4}V_1 \]
Thus, the wavelength corresponding to \(V_2\) is:
\[ \lambda_2 = \frac{hc}{eV_2} = \frac{hc}{e \cdot \frac{5}{4}V_1} = \frac{4hc}{5eV_1} \]

4. **Finding the change in wavelength**:
The change in wavelength \(\Delta \lambda\) is given by:
\[ \Delta \lambda = \lambda_1 - \lambda_2 \]
Substituting the expressions for \(\lambda_1\) and \(\lambda_2\):
\[ \Delta \lambda = \frac{hc}{eV_1} - \frac{4hc}{5eV_1} \]
Simplifying this expression:
\[ \Delta \lambda = \frac{hc}{eV_1} \left(1 - \frac{4}{5}\right) = \frac{hc}{eV_1} \cdot \frac{1}{5} = \frac{hc}{5eV_1} \]

5. **Solving for the initial voltage \(V_1\)**:
Rearranging the equation gives:
\[ V_1 = \frac{hc}{5e\Delta \lambda} \]

### Final Result:
The initial voltage applied to the tube is:
\[ V_1 = \frac{hc}{5e\Delta \lambda} \]