Question

De broglie postulated that the relationship = h/p is valid for relativistic particles. What is the de broglie wavelength for a (relativistic) electron having a kinetic energy of 3. 45 mev?

Answer 1

The de-broglie wavelength is 4.81×$10^{-18}$m.

We know that,

λ = $\frac{h}{p}$ and $E_{k}$ = $\frac{1}{2}mv^{2}$

by using the above two equations, we can deduce a relation between de-broglie wavelength and kinetic energy, that can be shown as,

λ = $\frac{h}{\sqrt{2M_{e} E_{k} } }$

λ = $\frac{6.6*10^{-34} }{\sqrt{2*9.1*10^{-31} _{} 3*3.45*10^{-3} _{} } }$

on solving the above equation

λ = $\frac{6.6*10^{-34} }{\sqrt{188.37*10^{-34} _{} } }$

λ = $\frac{6.6*10^{-34} }{13.72*10^{-17} }$

λ = $\frac{6.6*10^{-34} }{13.72*10^{-17} }$ = 4.81×$10^{-18}$m

Hence, the de-broglie wavelength is 4.81×$10^{-18}$m.

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