Apparatus:
The apparatus shown in Fig.1. It comprises (consists) of a glass tube G which can be evacuated through the side tube T. When the Ultraviolet (UV) light passes through a quartz window W and falls on an aluminum cathode C enclosed in G. An earthed metal screen A with a small central hole forms the anode. The cathode (C) can be continued (maintain) at a desired potential, positive or negative relative to the anode A. P1 and P2 are small metal electrodes connected to electrometers E1 and E2.
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| Fig.1 |
Working:
When C is raised to a negative potential and illuminated, negatively charged particles are produced and accelerated towards the anode. A few particles proceed(pass) through the hole in A and proceed with uniform velocity to P1. Their arrival at P1 is indicated by E1. By applying a uniform magnetic field B (represented by the dotted circle) perpendicular to the plane of the figure and directed towards the reader, the photoelectrons can be deflected towards P2. Their arrival at P2 is indicated by the deflection they produce in E2.
Variation of photoelectric current with cathode potential:
Lenard first studied the relation between current and the potential applied to C. When the cathode potential was several volts positive, the current was zero. When V were +2 volts, there was a feeble current (low current) showing that a few particles possessed enough velocity to overcome the retarding potential of 2 volts. When the potential was further decreased, the current increased and reached a saturation value for —20 volts. Fig.2 shows the variation of photoelectric current with cathode potential.
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| Fig.2 |
Determination of e/m:
After this preliminary study (investigation), Lenard applied to C a negative potential V, very huge(large)compared to the potential of 2 volts. The velocity imparted by the accelerating potential is so large that the velocity of the particles in the act of emission is negligible in comparison to it. Let be V -accelerating potential and v - velocity acquired by the photoelectrons.
Then,
½mv²= eV ... (1)
where,
e be the the charge and
m is the mass of the photoelectron.
Let R be the radius of the circular path reported (described) by the photoelectrons in the region of uniform magnetic field of strength B.
Then,
mv²/ R = Bev
v = BeR/m ...(2)
Substituting the value of v in the Equ (1),
½m(BeR/m)² = eV
Knowing V, B and R, e/m is calculated. Lenard found the value of e/m to be the same as that for electrons. This were certainly (clearly) shows that the photoparticles are nothing but electrons.
