Q.54 An electron is accelerated from rest through a potential difference of 200 V. The de
Broglie wavelength associated with this electron is _______ nm. (Rounded off to 2
decimal places) (Planck’s constant = 6.6 × 10–34 J s, 1eV=1.6 × 10–19 J, mass of an
electron = 9.1×10–31 kg)
Introduction
Discover how to compute the de Broglie wavelength of electron accelerated through 200 V potential difference using Planck’s constant (h = 6.6 × 10^{-34} J s), electron mass (m = 9.1 × 10^{-31} kg), and 1 eV = 1.6 × 10^{-19} J. This key CSIR NET Life Sciences/Physics problem tests wave-particle duality, yielding 0.09 nm rounded to two decimals.
Step-by-Step Solution
The electron gains kinetic energy KE = eV from the accelerating potential, where e = 1.6 × 10^{-19} J (since 1 eV = 1.6 × 10^{-19} J) and V = 200 V, yielding KE = 3.2 × 10^{-17} J.
Momentum follows from KE = \frac{p^2}{2m}, so p = \sqrt{2m \cdot KE}. With electron mass m = 9.1 × 10^{-31} kg, p = 7.63 × 10^{-24} kg m/s.
De Broglie wavelength is \lambda = \frac{h}{p}, where Planck’s constant h = 6.6 × 10^{-34} J s. Thus, \lambda = 8.65 × 10^{-11} m or 0.0865 nm, which rounds to 0.09 nm.
Verification Formula
A standard approximation confirms: \lambda \text{(in nm)} = \frac{1.226}{\sqrt{V}} for electrons in volts, giving \frac{1.226}{\sqrt{200}} = 0.0867 nm, matching the precise value before rounding.
Core Formula and Derivation
Kinetic energy gained: KE = eV = (1.6 × 10^{-19})(200) = 3.2 × 10^{-17} J.
Momentum: p = \sqrt{2mKE} = 7.63 × 10^{-24} kg m/s.
Wavelength: \lambda = \frac{h}{p} = 0.0865 nm ≈ 0.09 nm.
Approximation: \lambda = \frac{1.226}{\sqrt{V}} nm simplifies checks for de Broglie wavelength electron 200 V
Why No Options?
This numerical fill-in (CSIR NET style) has no multiple choices—pure calculation. Common errors: forgetting eV-to-J conversion or nm units (1 nm = 10^{-9} m).
Applications in Quantum Physics
Such wavelengths (~0.1 nm) match X-ray scales, vital for electron microscopy and diffraction experiments confirming de Broglie hypothesis.
