Electrons with de - Brogli wavelengtyh `lambda` fall on the target in
an X-ray tube.The cut off wavelength of the emitted Xrays is
(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 solve the problem of finding the cut-off wavelength of the emitted X-rays when electrons with a de Broglie wavelength \( \lambda \) fall on the target in an X-ray tube, we can follow these steps:

### Step 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.

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

### Step 3: Substitute Momentum into the de Broglie Equation
Substituting \( p \) into the de Broglie wavelength equation:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Step 4: Solve for Energy in Terms of Wavelength
Rearranging the above equation to express energy \( E \):
\[ E = \frac{h^2}{2m\lambda^2} \]

### Step 5: Relate Energy to Cut-off Wavelength
For X-ray emission, the energy \( E \) can also be expressed in terms of the cut-off wavelength \( \lambda_0 \):
\[ E = \frac{hc}{\lambda_0} \]
where \( c \) is the speed of light.

### Step 6: Set the Two Energy Equations Equal
Setting the two expressions for energy equal to each other:
\[ \frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2} \]

### Step 7: Solve for the Cut-off Wavelength \( \lambda_0 \)
Cross-multiplying gives:
\[ hc \cdot 2m\lambda^2 = h^2 \lambda_0 \]
Dividing both sides by \( h \):
\[ 2mc\lambda^2 = h\lambda_0 \]
Finally, solving for \( \lambda_0 \):
\[ \lambda_0 = \frac{2mc\lambda^2}{h} \]

### Conclusion
The cut-off wavelength \( \lambda_0 \) of the emitted X-rays is given by:
\[ \lambda_0 = \frac{2mc\lambda^2}{h} \]

### Answer
The correct option is (a) \( \lambda_0 = \frac{2mc\lambda^2}{h} \).

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