What is the de Broglie wavelength of an electron with kinetic energy of 81 eV?
Answer & explanation
Correct answer: option 1
The de Broglie wavelength is given by
\(\lambda \) =\(\frac{h}{\sqrt {2mK}}\)
Kinetic energy, K = 81 eV = 81×1.6×10-19 J
\(\lambda \) = \(\frac{6.626 × 10^{-34}}{\sqrt{2 × 9.1 × 10^{-31} × 81 × 1.6 × 10^{-19}}}\)
= 0.136 nm