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. `lambda_(0)=(2mclambda^(2))/(h)`
B. `lambda_(0)=(2h)/(mc)`
C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/(h^(2))`
D. `lambda_(0)=lambda`

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 formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

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

3. **Express Kinetic Energy in Terms of Wavelength**:
From the de-Broglie relation, we can express momentum as:
\[ p = \frac{h}{\lambda} \]
Substituting this into the kinetic energy formula gives:
\[ KE = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2} \]

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

5. **Equate Kinetic Energy to X-ray Energy**:
At the cut-off wavelength, the entire kinetic energy of the electron is converted into the energy of the emitted X-ray:
\[ \frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_0} \]

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

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