Electrons with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

Options

A. `lamda_(0)=(2mclamda^(2))/(h)`
B. `lamda_(0)=(2h)/(mc)`
C. `lamda_(0)=(2m^(2)c^(2)lamda^(2))/(h^(2))`
D. `lamda_(0)=lamda`

Explanation

To find the cut-off wavelength of the emitted X-ray when electrons with a de-Broglie wavelength \( \lambda \) fall on the target in an X-ray tube, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de-Broglie Wavelength**:
The de-Broglie wavelength \( \lambda \) of an electron is given by the equation:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

2. **Relate Momentum to Energy**:
The momentum \( p \) of the electron can also be expressed in terms of its kinetic energy \( E \):
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the electron.

3. **Substituting Momentum into the de-Broglie Equation**:
Substituting the expression for momentum into the de-Broglie wavelength equation gives:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

4. **Energy of the Emitted X-ray**:
The energy \( E \) of the emitted X-ray can be expressed in terms of its cut-off wavelength \( \lambda_0 \):
\[ E = \frac{hc}{\lambda_0} \]
where \( c \) is the speed of light.

5. **Substituting X-ray Energy into the Wavelength Equation**:
Now, substitute the expression for \( E \) into the de-Broglie wavelength equation:
\[ \lambda = \frac{h}{\sqrt{2m \left(\frac{hc}{\lambda_0}\right)}} \]

6. **Simplifying the Equation**:
Rearranging the equation gives:
\[ \lambda = \frac{h \lambda_0^{1/2}}{\sqrt{2m \cdot hc}} \]
Squaring both sides results in:
\[ \lambda^2 = \frac{h^2 \lambda_0}{2m \cdot hc} \]

7. **Solving for the Cut-off Wavelength \( \lambda_0 \)**:
Rearranging the above equation to solve for \( \lambda_0 \):
\[ \lambda_0 = \frac{2m c \lambda^2}{h} \]

### Final Result:
The cut-off wavelength \( \lambda_0 \) of the emitted X-ray is given by:
\[ \lambda_0 = \frac{2m c \lambda^2}{h} \]