Question

Electrons used in an electron microscope are accelerated by a voltage of $25\, kV$. If the voltage is increased to $100\, kV$ then the de-Broglie wavelength associated with the electrons would

• A. increase by 2 times
• B. decrease by 2 times
• C. decrease by 4 times
• D. increase by 4 times

Answer 1

The Correct Option is B

To solve this problem, we need to find out the relationship between the accelerating voltage of electrons and the de-Broglie wavelength associated with them. The de-Broglie wavelength \( \lambda \) is given by the equation:

\lambda = \frac{h}{p}

where:

• h is Planck's constant.
• p is the momentum of the electron.

The momentum \( p \) of an electron accelerated by a voltage \( V \) can be found using:

p = \sqrt{2meV}

where:

• m is the mass of the electron.
• e is the charge of the electron.
• V is the accelerating voltage.

Substituting for \( p \) in the de-Broglie wavelength formula:

\lambda = \frac{h}{\sqrt{2meV}}

This shows that the de-Broglie wavelength is inversely proportional to the square root of the voltage \( V \). Therefore, as the voltage increases, the wavelength decreases. Specifically, if the voltage is increased from 25 kV to 100 kV, we have:

\lambda \propto \frac{1}{\sqrt{V}}

If the initial voltage is \( V_1 = 25 \, \text{kV} \) and the final voltage is \( V_2 = 100 \, \text{kV} \), then the initial and final wavelengths \( \lambda_1 \) and \( \lambda_2 \) are related by:

\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{25}{100}} = \frac{1}{2}

Thus, the final wavelength \( \lambda_2 \) becomes half of the initial wavelength \( \lambda_1 \), meaning the de-Broglie wavelength decreases by 2 times.

Therefore, out of the given options, the correct answer is that the de-Broglie wavelength associated with the electrons would "decrease by 2 times".