Electron beam used in an electron microscope. When accelerated by a voltage of 20 kV, has a de-Broglie wavelength of `lambda_0`. If the voltage is increased 40kV, then the de-Broglie wavelength associated with the electron beam would be :

Options

A. `(lambda_0)/2`
B. `3lambda_0`
C. `lambda_0/sqrt2`
D. `lambda_0`

Answer

To solve the problem, we need to understand the relationship between the de-Broglie wavelength of an electron and the accelerating voltage. The de-Broglie wavelength \(\lambda\) of a particle is given by the formula:

\[ \lambda = \frac{h}{p} \]

where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

For an electron accelerated through a potential difference \(V\), the kinetic energy gained by the electron is given by:

\[ KE = eV \]

where \(e\) is the charge of the electron. The kinetic energy can also be expressed in terms of momentum:

\[ KE = \frac{p^2}{2m} \]

where \(m\) is the mass of the electron. Setting these two expressions for kinetic energy equal gives us:

\[ eV = \frac{p^2}{2m} \]

From this, we can solve for momentum \(p\):

\[ p = \sqrt{2m eV} \]

Substituting this expression for momentum back into the de-Broglie wavelength formula gives:

\[ \lambda = \frac{h}{\sqrt{2m eV}} \]

Now, we will denote the de-Broglie wavelength at 20 kV as \(\lambda_0\):

\[ \lambda_0 = \frac{h}{\sqrt{2m e \cdot 20000}} \quad \text{(since 20 kV = 20000 V)} \]

When the voltage is increased to 40 kV, the new de-Broglie wavelength \(\lambda_1\) becomes:

\[ \lambda_1 = \frac{h}{\sqrt{2m e \cdot 40000}} \quad \text{(since 40 kV = 40000 V)} \]

Now, we can find the ratio of the two wavelengths:

\[ \frac{\lambda_0}{\lambda_1} = \frac{\sqrt{2m e \cdot 40000}}{\sqrt{2m e \cdot 20000}} \]

This simplifies to:

\[ \frac{\lambda_0}{\lambda_1} = \sqrt{\frac{40000}{20000}} = \sqrt{2} \]

Thus, we can express \(\lambda_1\) in terms of \(\lambda_0\):

\[ \lambda_1 = \frac{\lambda_0}{\sqrt{2}} \]

Therefore, when the voltage is increased from 20 kV to 40 kV, the de-Broglie wavelength associated with the electron beam becomes:

\[ \lambda_1 = \frac{\lambda_0}{\sqrt{2}} \]

### Final Answer:
\(\lambda_1 = \frac{\lambda_0}{\sqrt{2}}\)