Question
An electron has a de Broglie wavelength of $2.80 \times 10^{-10} \mathrm{m}$ . Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).
Step 1
The de Broglie wavelength (\(\lambda\)) is related to the momentum (\(p\)) of a particle by the formula: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg/s}\)). Show more…
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