Question:medium

The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of 6600 Å is nearly (Planck's constant = \(6.6 \times 10^{-34}\) Js and mass of electron = \(9 \times 10^{-31}\)kg)

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There are two key formulas for kinetic energy. The classical one is \(K.E. = \frac{1}{2}mv^2\). The one in terms of momentum, \(K.E. = p^2/2m\), is often more useful in quantum physics problems where the de Broglie wavelength (which gives momentum) is known.

Updated On: Mar 30, 2026

\(5.56 \times 10^{-25}\) eV
1.88 eV
\(5.56 \times 10^{-25}\) J
1.88 J

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The Correct Option is C

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