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: mass of electron = 9.1 $\times 10$ ⁻³¹ kg, charge = 1.6 $\times 10$ ⁻¹⁹ C, Planck's constant = 6.63 $\times 10$ ⁻³⁴ J-s)

0.123 nm

1.23 nm

0.0123 nm

12.3 nm

Correct Answer

⚙️ TE • Technical Concept & PrincipleGeneral SciencePhysics: Electricity

Option A

0.123 nm

Quick Summary:

The de Broglie wavelength of an accelerated electron is inversely proportional to the square root of the applied potential difference. For an electron accelerated through $100\text{ V}$ , the wavelength is calculated as approximately $0.123\text{ nm}$ .

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🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of light (a wave) behaving like a particle in certain cases; the de Broglie wavelength represents the dual nature where a tiny particle like an electron acts as a wave, similar to how sound waves ripple through the air.

🔢 Key Formulas

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

$\lambda \approx \frac{1.227}{\sqrt{V}}\text{ nm}$ — simplified shortcut for electrons

⚙️ Working Principle

According to de Broglie's hypothesis, moving particles exhibit wave-like properties with a wavelength $\lambda = \frac{h}{p}$ . Since the kinetic energy of an electron is $K = eV = \frac{p^2}{2m}$ , we derive $\lambda = \frac{h}{\sqrt{2meV}}$ . Substituting the constants, we get the simplified formula $\lambda \approx \frac{1.227}{\sqrt{V}}\text{ nm}$ .

📌 Key Points

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Wave-particle duality applies to all matter, but it is only significant for particles with extremely small masses like electrons.
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Higher accelerating potential results in a smaller wavelength, allowing higher resolution in electron microscopy.
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Planck's constant ( $h$ ) relates the energy of particles to the frequency of their wave-like counterparts.

✅ Advantages

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Allows for high-resolution imaging in transmission electron microscopes (TEM).

❌ Disadvantages / Limitations

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The wave nature is negligible for macroscopic objects due to their large mass relative to $h$ .

🛠️ Applications / Uses

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Electron microscopy
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Quantum mechanical modeling of atoms

📄 Additional Information

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The constant $1.227$ comes from $\frac{h}{\sqrt{2me}}$ .
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Option B ( $1.23\text{ nm}$ ) corresponds to an accelerating potential of $1\text{ V}$ .

📊 Diagram / Illustration

✅

A is correct — The calculation $\lambda = \frac{1.227}{\sqrt{100}} = 0.1227\text{ nm} \approx 0.123\text{ nm}$ confirms the result.

Core Concepts Used

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de Broglie Hypothesis Wave-Particle Duality Photoelectric Effect related Physics

💡 EXAM TIP

Always remember the shortcut $\lambda \approx \frac{12.27}{\sqrt{V}}\text{ \AA}$ (Angstroms), where $1\text{ \AA} = 0.1\text{ nm}$ to solve these problems in seconds.

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