Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Explanation

KEU IDEAS
(1) We can find the electron.s de Broglie wavelength `lambda` from Eq. 37-9 `(lambda=h//p)` if we first find the magnitude of its momentum p. (2) We find p from the given kinetic energy K of the electron. That kinetic energy is much less than the rest energy of an electron (0.511 MeV, from Table 36-3). Thus, we can get by with the classical approximations for momentum `p(=mv)` and kinetic energy `K(=(1//2)mv^(2))`.
Calculations : We are given the value of the kinetic energy. So, in order to use the de Broglie relation, we first solve the kinetic energy quation for v and then substitute into the monentum equation, finding
`p=sqrt(2mk)`
`=sqrt((2)(9.11xx10^(-31)kg)(120eV)(1.60xx10^(-19)J//eV))`
`=5.91xx10^(-24)kg.m//s`
From Eq. 37.9 then
`lambda=(h)/(p)`
`=(6.63xx10^(-34)J.s)/(5.91xx10^(-24)kg.m//s)`
`=1.12xx10^(-10)mm=112 p m`
This wavelength associated with the electron is about the size of a typical atom. If we increase the electron.s kinetic energy, the wavelength becomes even smaller.