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

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

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

where:

• \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)).
• \(m\) is the electron mass (\(9.109 \times 10^{-31} \, \text{kg}\)).
• \(e\) is the elementary charge (\(1.602 \times 10^{-19} \, \text{C}\)).
• \(V\) is the accelerating voltage in volts.

Initially, the electron is accelerated by a voltage of \(20 \, \text{kV}\), leading to a de-Broglie wavelength of \(\lambda_0\). When the voltage is increased to \(40 \, \text{kV}\), we can use the same formula to find the new wavelength \(\lambda'\).

For the initial voltage \(V_1 = 20 \, \text{kV}\), the wavelength is:

\(\lambda_0 = \frac{h}{\sqrt{2meV_1}}\)

For the new voltage \(V_2 = 40 \, \text{kV}\), the wavelength \(\lambda'\) is:

\(\lambda' = \frac{h}{\sqrt{2meV_2}}\)

Dividing the two equations:

\(\frac{\lambda'}{\lambda_0} = \frac{\sqrt{2meV_1}}{\sqrt{2meV_2}}\)

\(\frac{\lambda'}{\lambda_0} = \frac{\sqrt{V_1}}{\sqrt{V_2}}\)

Substitute the voltages:

\(\frac{\lambda'}{\lambda_0} = \frac{\sqrt{20}}{\sqrt{40}} = \frac{\sqrt{20}}{\sqrt{2 \times 20}} = \frac{1}{\sqrt{2}}\)

Hence, the new wavelength \(\lambda'\) is:

\(\lambda' = \frac{\lambda_0}{\sqrt{2}}\)

Therefore, the correct answer is \(\frac{\lambda_0}{\sqrt{2}}\).