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

Answer

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

### Step 1: Understand the relationship between momentum and de-Broglie wavelength
The momentum \( p \) of an electron can be expressed in terms of its de-Broglie wavelength \( \lambda \):
\[ p = \frac{h}{\lambda} \]
where \( h \) is Planck's constant.

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

### Step 3: Substitute momentum into the kinetic energy equation
Substituting the expression for momentum into the kinetic energy equation gives:
\[ KE = \frac{1}{2m} \left(\frac{h}{\lambda}\right)^2 = \frac{h^2}{2m\lambda^2} \]

### Step 4: Relate kinetic energy to the emitted X-ray wavelength
The energy of a photon emitted (cut-off wavelength \( \lambda_0 \)) can be expressed as:
\[ E = \frac{hc}{\lambda_0} \]
where \( c \) is the speed of light.

### Step 5: Set the kinetic energy equal to the photon energy
Assuming all the kinetic energy of the electron is converted into the energy of the emitted X-ray photon, we can set the two expressions for energy equal to each other:
\[ \frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_0} \]

### Step 6: Solve for the cut-off wavelength \( \lambda_0 \)
Rearranging the equation to solve for \( \lambda_0 \):
\[ \lambda_0 = \frac{2m c \lambda^2}{h} \]

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