Calculate the de Broglie wavelength of an electron accelerated... | General Science MCQ | Examoogle

General SciencePhysics: Electricity

Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V. (Given: h = 6.63 $\times 10$ ⁻³⁴ Js, m = 9.1 $\times 10$ ⁻³¹ kg, e = 1.6 $\times 10$ ⁻¹⁹ C)

0.123 nm

0.246 nm

0.061 nm

0.184 nm

Correct Answer

⚙️ TE • Technical Concept & PrincipleGeneral SciencePhysics: Electricity

Option A

0.123 nm

Quick Summary:

The de Broglie wavelength of an electron accelerated through a potential difference is determined by its kinetic energy gained from the electric field. Using the derived formula $\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$ , substituting $V = 100$ V gives $\lambda \approx 0.123$ nm.

🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of the electron as a wave-like surfer; the higher the voltage (the 'tide' or 'acceleration'), the faster the surfer goes, and the shorter the distance between the crests of the waves (wavelength) becomes.

🧠 Memory Mnemonic / Shortcut Aid

Remember '12.27 over root V' (in Angstroms) for quick calculations.

🔢 Key Formulas

$\lambda = \frac{h}{\sqrt{2meV}}$ — General de Broglie wavelength formula for charged particles

$\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$ — Simplified formula for electrons

⚙️ Working Principle

When an electron is accelerated through a potential difference $V$ , it gains kinetic energy $K = eV$ . According to de Broglie's hypothesis, the wavelength $\lambda$ associated with a particle of mass $m$ and momentum $p$ is $\lambda = \frac{h}{p}$ . Since $K = \frac{p^2}{2m}$ , we have $p = \sqrt{2mK} = \sqrt{2meV}$ . Substituting this into the wavelength equation yields $\lambda = \frac{h}{\sqrt{2meV}}$ .

📌 Key Points

▸
Wavelength is inversely proportional to the square root of the accelerating potential.
▸
The wave-particle duality explains why electrons exhibit diffraction patterns similar to light.
▸
This calculation assumes non-relativistic conditions, which holds for $V = 100$ V.

✅ Advantages

▸
Provides fundamental insight into matter waves.
▸
Basis for Electron Microscopy technologies.

❌ Disadvantages / Limitations

▸
Does not account for relativistic mass increase at very high potentials.
▸
Neglects thermal energy of the electron before acceleration.

🛠️ Applications / Uses

▸
Transmission Electron Microscopy (TEM)
▸
Quantum mechanical modeling of semiconductor devices

📄 Additional Information

▸
Standard value: $h \approx 6.63 \times 10^{-34} \text{ Js}$ .
▸
Option B (0.246 nm) is obtained if one forgets the square root of V; Option C (0.061 nm) may arise from miscalculating the constant or the square root factor.

📊 Diagram / Illustration

✅

A is correct — The calculated de Broglie wavelength for an electron at 100 V is 0.123 nm.

Core Concepts Used

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De Broglie Hypothesis Photoelectric/Acceleration Physics Matter Waves

💡 EXAM TIP

For competitive exams, memorize $\lambda = \frac{12.27}{\sqrt{V}} \mathring{A}$ for electrons to save time, where $\mathring{A} = 10^{-10} \text{ m}$ .

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