1.

An electron is accelerated by a potential difference of 10^(2) V Find the group and the phase velocities of the de Broglie waves. Do the same for a potential difference of 10^(6) V.

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Solution :(1) At the accelerating potential `Phi_(1)=10^(2)V` the electron is a nonrelativistic particle. Its momentum is `p_(1)=sqrt(2mePhi_(1))` and the velocity is `v_(1)=sqrt(2ePhi_(1)//m)`. But the group velocity of the de BROGLIE wave is equal to the particle.s velocity: `U_(1)=v_(1)`. The phase velocity is
`u_(1)=e^(2)/v_(1)=e^(2)sqrt(m/(2ePhi_(1)))`
(2) At the accelerating potential `Phi_(2)=10^(5)V` the electron is a relativistic particle. Its momentum is found from the condition `sqrt(epsilon_(2)^(2)+p^(2)c^(2))=epsi_(0)+ePhi_(2)`, from which is follows that `p_(2)=1/csqrt(ePhi_(2)(2epsi_(0)+ePhi_(2)))`. The mass is found from the condition `m_(2)c^(2)=epsi_(0)+ePhi_(2)`, where `m_(2)=1/c^(2)(epsi_(0)+ePhi_(2))`. The group velocity is
`U_(2)=v_(2)=p_(2)/m_(2)=(csqrt(ePhi_(2)(2epsi_(0)+ePhi_(2))))/(epsi_(0)+ePhi_(2))`
The phase velocity is
`u_(2)=c^(2)/v_(2)=(c(epsi_(0)+ePhi_(2)))/(sqrt(ePhi_(2)(2epsi_(0)+ePhi_(2))))`
Note that in this case the energy is conveniently expressed not in JOULES, but in kiloelectron-volts, since `epsi_(0)=520keV`.


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