Question

In an electron microscope, accelerated electrons have wavelength of $0.011$ nm. Calculate the voltage through which electrons were accelerated to attain this wavelength.

(Take $e = 1.6 \times 10^{-19}$ C, $m_e = 9 \times 10^{-31}$ kg, $h = 6.6 \times 10^{-34}$ J.s)

Hint:

Use \(V = \dfrac{h^2}{2m_e e\lambda^2}\) with \(\lambda = 1.1 \times 10^{-11}\) m. The result is about 12,500 V.

Answer 1

Step 1: Understanding the Concept:
An electron accelerated from rest through a potential difference \(V\) gains kinetic energy \(eV\). Its de Broglie wavelength is \(\lambda = \dfrac{h}{p}\), where \(p\) is its momentum.

Step 2: Key Formula:
\[ p = \sqrt{2m_e eV} \quad \Rightarrow \quad \lambda = \frac{h}{\sqrt{2m_e eV}} \]
Solving for \(V\):
\[ V = \frac{h^2}{2m_e e \lambda^2} \]

Step 3: Substitute the Values:
Given \(\lambda = 0.011\ \text{nm} = 1.1 \times 10^{-11}\) m, \(h = 6.6 \times 10^{-34}\) J s, \(m_e = 9 \times 10^{-31}\) kg, \(e = 1.6 \times 10^{-19}\) C.
\[ V = \frac{(6.6 \times 10^{-34})^2}{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times (1.1 \times 10^{-11})^2} \]
Numerator: \(4.356 \times 10^{-67}\). Denominator: \(2.88 \times 10^{-49} \times 1.21 \times 10^{-22} = 3.485 \times 10^{-71}\).

Step 4: Calculate:
\[ V = \frac{4.356 \times 10^{-67}}{3.485 \times 10^{-71}} = 1.25 \times 10^{4}\ \text{V} \]

Final Answer:
The electrons were accelerated through about 12,500 V (12.5 kV). \[ \boxed{V = 1.25 \times 10^{4}\ \text{V}} \]