JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves J.J. Thomson's Experiment

(1) It's working is based on the fact that if a beam of electron is subjected to the crossed electric field \[\overrightarrow{E}\] and magnetic field \[\overrightarrow{B}\], it experiences a force due to each field. In case the forces on the electrons in the electron beam due to these fields are equal and opposite, the beam remains undeflected.

(2) When no field is applied, the electron beam produces illuminations at point P.

(3) In the presence of any field (electric and magnetic) electron beam deflected up or down (illumination at \[P'\] or \[P''\])

(4) If both the fields are applied simultaneously and adjusted such that electron beam passes undeflected and produces illumination at point P.

In this case; Electric force = Magnetic force

\[\Rightarrow \] eE = evB  \[\Rightarrow \] \[v=\frac{E}{B};\] v = velocity of electron

(5) As electron beam accelerated from cathode to anode its loss in potential energy appears as gain in the K.E. at the anode. If suppose V is the potential difference between cathode and anode then, loss in potential energy = eV

And gain in kinetic energy at anode will be K.E. \[=\frac{1}{2}m{{v}^{2}}\] i.e.

\[eV=\frac{1}{2}m{{v}^{2}}\]\[\Rightarrow \]\[\frac{e}{m}=\frac{{{v}^{2}}}{2V}\]\[\Rightarrow \]\[\frac{e}{m}=\frac{{{E}^{2}}}{2V{{B}^{2}}}\]

Thomson found, \[\frac{e}{m}=1.77\times {{10}^{11}}C/kg.\]

If one includes the relativistic variation of mass with speed \[(m={{m}_{0}}/\sqrt{1-{{v}^{2}}/{{c}^{2}}})\], then specific charge of an electron decreases with the increase in its velocity.

(6) The deflection of an electron in a purely electric field is given by \[y=\frac{1}{2}\left( \frac{eE}{m} \right).\frac{{{l}^{2}}}{{{v}^{2}}}\]; where l = Length of each plate, y = deflection of electron in the field region, v = speed of the electron.