A potential difference of `10^(3)` V is applied across an X-ray tube. The ratio of the de-Broglie wavelength of X-rays produced is

Options

A. `1/20`
B. `1//100`
C. 1
D. `1//10^(4)`

Answer

To solve the problem of finding the ratio of the de-Broglie wavelength of X-rays produced when a potential difference of \(10^3\) V is applied across an X-ray tube, we can follow these steps:

### Step 1: Understand the Wavelength of X-rays
The wavelength of X-rays can be derived from the energy of the X-rays produced. The energy \(E\) of the X-rays can be expressed in terms of the potential difference \(V\) applied across the tube:
\[ E = eV \]
where \(e\) is the charge of an electron (approximately \(1.6 \times 10^{-19}\) C).

The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)) and \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)). Rearranging this gives:
\[ \lambda = \frac{hc}{eV} \]
Let this be Equation (1).

### Step 2: Understand the De-Broglie Wavelength
The de-Broglie wavelength \(\lambda'\) of a particle can be expressed as:
\[ \lambda' = \frac{h}{p} \]
where \(p\) is the momentum of the particle. For an electron accelerated through a potential difference \(V\), the kinetic energy \(E\) is given by:
\[ E = \frac{p^2}{2m} \]
This can be rearranged to find momentum:
\[ p = \sqrt{2mE} \]
Substituting \(E = eV\) gives:
\[ p = \sqrt{2meV} \]
Thus, the de-Broglie wavelength becomes:
\[ \lambda' = \frac{h}{\sqrt{2meV}} \]
Let this be Equation (2).

### Step 3: Find the Ratio of the Wavelengths
We need to find the ratio of the wavelength of X-rays to the de-Broglie wavelength:
\[ \text{Ratio} = \frac{\lambda}{\lambda'} = \frac{\frac{hc}{eV}}{\frac{h}{\sqrt{2meV}}} \]
This simplifies to:
\[ \text{Ratio} = \frac{hc}{eV} \cdot \frac{\sqrt{2meV}}{h} = \frac{c \sqrt{2m}}{e} \]

### Step 4: Substitute Known Values
Now we substitute the known values:
- \(c = 3 \times 10^8 \, \text{m/s}\)
- \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)
- \(V = 10^3 \, \text{V}\)

Calculating the ratio:
\[ \text{Ratio} = \frac{3 \times 10^8 \cdot \sqrt{2 \cdot 9.1 \times 10^{-31}}}{1.6 \times 10^{-19}} \]

### Step 5: Calculate the Numerical Value
Calculating the square root and the final value:
\[ \sqrt{2 \cdot 9.1 \times 10^{-31}} \approx 1.34 \times 10^{-15} \]
Now substituting back:
\[ \text{Ratio} \approx \frac{3 \times 10^8 \cdot 1.34 \times 10^{-15}}{1.6 \times 10^{-19}} \approx \frac{4.02 \times 10^{-7}}{1.6 \times 10^{-19}} \approx 1.009 \]

### Conclusion
Thus, the ratio of the de-Broglie wavelength of X-rays produced is approximately \(1\), which corresponds to option C.

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