|q⟩ Bad Qubits

← Question Bank

Multiple choice
An electron (m=9.109e31m = 9.109e-31 kg) is accelerated through V=54V = 54 V (the Davisson-Germer energy), so p=2meVp = \sqrt{2 m e V} with e=1.602e19e = 1.602e-19 C. Its de Broglie wavelength is λ=h/p\lambda = h/p with h=6.626e34h = 6.626e-34 J*s. What is λ\lambda in picometers (11 pm =1e12= 1e-12 m)?