Question

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 to $40 kV$, then the de-Broglie wavelength associated with the electron beam would be:

• A. $\frac{\lambda_0}{2}$
• B. $3 \lambda_0$
• C. $\frac{\lambda_0}{\sqrt{2}}$
• D. $9 \lambda_0$

Answer 1

The Correct Option is C

1. The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{\sqrt{2meV}}, \] where \(h\) is Planck's constant, \(m\) is the mass of the electron, \(e\) is the charge of the electron, and \(V\) is the accelerating voltage.
2. The wavelength is inversely proportional to the square root of the voltage: \[ \lambda \propto \frac{1}{\sqrt{V}}. \]
3. For \(V = 20 \, \text{kV}\), \(\lambda = \lambda_0\). For \(V = 40 \, \text{kV}\): \[ \lambda = \lambda_0 \times \frac{\sqrt{20}}{\sqrt{40}} = \frac{\lambda_0}{\sqrt{2}}. \]
Thus, the de-Broglie wavelength is \(\frac{\lambda_0}{\sqrt{2}}\). The de-Broglie wavelength is inversely proportional to the square root of the accelerating voltage. Doubling the voltage reduces the wavelength by a factor of \(\sqrt{2}\).