Question

An electron, starting from rest, accelerates through a potential difference of 25.0 V. What is the final de Broglie wavelength of the electron, assuming that its final speed is not relativistic

Answer 1

Answer: λ = 0.245 nm

Explanation:

KE = qV

V = 25.0 V

where q = e = 1.6*10^-19

KE = (1.6*10^-19)(25.0 V)

KE= 4.0 * 10^-18 J

λ = $\frac{h}{\sqrt{2mKE} }$

mass of electron = 9.11 * 10^-34

λ = $\frac{6.626*10^{-34} }{\sqrt{(2)(9.11*10^{-31})(4.0*10^{-18}) } }$

λ = 2.45 e -10 m

λ = 0.245 nm