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 step by step, we will analyze the relationship between the voltage applied to the Coolidge X-ray tube and the wavelength of the emitted X-rays.

### Step 1: Understand the relationship between energy, voltage, and wavelength.
The energy of a photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength.

The energy of the photon is also related to the voltage \( V \) applied to the X-ray tube:
\[ E = eV \]
where \( e \) is the charge of an electron.

### Step 2: Set up the equations for initial and final conditions.
Let the initial voltage be \( V \) and the initial wavelength be \( \lambda \). The energy of the photon at this voltage is:
\[ E = eV = \frac{hc}{\lambda} \]

Now, if the voltage is increased by 25%, the new voltage \( V' \) is:
\[ V' = 1.25V = \frac{5}{4}V \]

The new wavelength \( \lambda' \) corresponding to this new voltage can be expressed as:
\[ E' = eV' = \frac{hc}{\lambda'} \]

### Step 3: Relate the initial and final wavelengths.
From the equations for energy, we have:
\[ eV = \frac{hc}{\lambda} \quad \text{(1)} \]
\[ eV' = \frac{hc}{\lambda'} \quad \text{(2)} \]

Dividing equation (1) by equation (2):
\[ \frac{\lambda'}{\lambda} = \frac{V}{V'} \]

Substituting \( V' = \frac{5}{4}V \):
\[ \frac{\lambda'}{\lambda} = \frac{V}{\frac{5}{4}V} = \frac{4}{5} \]

This implies:
\[ \lambda' = \frac{4}{5} \lambda \]

### Step 4: Determine the shift in wavelength.
The shift in wavelength \( \Delta \lambda \) can be defined as:
\[ \Delta \lambda = \lambda - \lambda' \]

Substituting \( \lambda' \):
\[ \Delta \lambda = \lambda - \frac{4}{5} \lambda = \frac{1}{5} \lambda \]

### Step 5: Express the initial voltage in terms of the shift.
From the relationship \( \Delta \lambda = \frac{1}{5} \lambda \), we can express \( \lambda \) as:
\[ \lambda = 5 \Delta \lambda \]

Now, substituting \( \lambda \) back into the energy equation:
\[ eV = \frac{hc}{\lambda} = \frac{hc}{5 \Delta \lambda} \]

Thus, we can express the voltage \( V \) as:
\[ V = \frac{hc}{5e \Delta \lambda} \]

### Conclusion
The initial voltage applied to the tube can be expressed as:
\[ V = \frac{hc}{5e \Delta \lambda} \]