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Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V. (Given: mass of electron = 9.1 ×10⁻³¹ kg, charge = 1.6 ×10⁻¹⁹ C, Planck's constant = 6.63 ×10⁻³⁴ J-s)
0.123 nm
1.23 nm
0.0123 nm
12.3 nm
0.123 nm
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 V, the wavelength is calculated as approximately 0.123 nm.
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 V, the wavelength is calculated as approximately 0.123 nm.
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.
λ=2meVh — de Broglie wavelength formula
λ≈V1.227 nm — simplified shortcut for electrons
According to de Broglie's hypothesis, moving particles exhibit wave-like properties with a wavelength λ=ph. Since the kinetic energy of an electron is K=eV=2mp2, we derive λ=2meVh. Substituting the constants, we get the simplified formula λ≈V1.227 nm.
Wave-particle duality applies to all matter, but it is only significant for particles with extremely small masses like electrons.
Higher accelerating potential results in a smaller wavelength, allowing higher resolution in electron microscopy.
Planck's constant (h) relates the energy of particles to the frequency of their wave-like counterparts.
Allows for high-resolution imaging in transmission electron microscopes (TEM).
The wave nature is negligible for macroscopic objects due to their large mass relative to h.
Electron microscopy
Quantum mechanical modeling of atoms
The constant 1.227 comes from 2meh.
Option B (1.23 nm) corresponds to an accelerating potential of 1 V.
A is correct — The calculation λ=1001.227=0.1227 nm≈0.123 nm confirms the result.
Always remember the shortcut λ≈V12.27 A˚ (Angstroms), where 1 A˚=0.1 nm to solve these problems in seconds.