Since, electron is accelerated through a potential difference
V, its initial velocity
v0 is given by
mv02=eV Or
v0=√−−−−−−−−−−−−−(1)
Since, initial velocity is parallel to plates or normal to the direction of electric field, component of velocity parallel to plates remains constant as
v0.
Hence, time taken by the electron to cross electric field is
t0= Now consider motion of electron, normal to plates.
At some instant
t, its acceleration
=== Let velocity component normal to plates be
vy ∴vy= ∴vydv=t0tdt or
vy=t02=−−−−−−−−−−(2) If . is angular deviation of electron from its initial direction of motion, then pitch of its helical path.
p=vcos(90−θ) ∴p=vsinθ=vy or
p= p=9mm