Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV.

Calculate the de Broglie wavelength of an electron that has been accel

1.	Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV.
Answer» Solution :`gamma =h/(mv)` Potendial difference of an ELECTRON=V=1keV. `THEREFORE "Potential energy "=1/2mv^(2)=eV` e=charge of an electron=`1.609xx10^(-19)c` `kv=1000V` `therefore "Potential energy"=1.609xx10^(-19)xx1000` `=1.609xx10^(-16)` `1/2mv^(2)=1.609xx10^(16)V``m=9.1xx10^(-31)kg` `gamma=h/(mv)` `v^(2)=(2xx1.609xx10^(-16))/(9.1xx10^(-31))` `v=sqrt(2xx1.609xx10^(-16))/(9.6xx10^(-31)` =`5.9xx10^(7)ms^(-1)` `gamma=h/(mv)=(6.626xx10^(-34))/(9.1xx10^(-31)xx5.93xx10^(7))` `=1.2xx10^(11)m` `gamma=1.2xx10^(-11)m`