34 The Electron Microscope 



tion. The difference is only apparent, as can be seen from a 

 simple consideration. 



• Figure 13 shows a strongly magnified picture of the electron 

 gun and a few characteristic electron trajectories. In order to 

 make the drawing clearer, the cathode is assumed flat as in 

 cathode ray tubes instead of hairpin-shaped. This makes prac- 

 tically little difference, as the active area of the cathode is usually 

 only a few tenths of a mm in diameter. Electrons will start from 

 every point of this area with different initial velocities and in 

 different directions. Three of these trajectories are shown in 

 the ilustration for three selected points, one starting at right 

 angles to the cathode, the other two tangentially. 



The accelerating field penetrates into the aperture of the shield 

 and produces in it a very powerful lens. This bends the trajec- 

 tories which have started at right angles to the cathode strongly 

 ^toward the axis which they cross at a distance of usually 1 mm or 

 less from the cathode. This point would be the image of the 

 cathode, if the electrons all started at right angles to it, which 

 would be the case if they had no initial velocities. Because of 

 the tangential velocities the cross-over w411 be a disk. Strictly 

 speaking, this disk has no well-defined limits. The intensity 

 distribution in it is a reproduction of the Maxwellian velocity 

 distribution of the electrons, and falls off with the radius r ac- 



cording to a law exponent 



[ 



a 



The constant, a, can 



be taken to represent the equivalent diameter of the disk. The 

 cross-over was first discovered and thoroughly discussed by 

 Maloff and Epstein.^^ 



Outside the electron gun the trajectories pass through the 

 condenser lens, and are bent parallel to the axis. (CoUimated.) 

 Only the principal trajectories, which have started at right 

 angles to the cathode, can be exactly collimated. There remains 

 a certain divergence in the beam which is illustrated by the 

 angle 8. The root mean square value of it can be calculated as 

 follows : The mean square tangential velocity components at the 

 cathode are hnv^ = hnVy^ = IkT, therefore, the mean square 

 tangential velocity is \niVt' = hn{v:^- -\- Vy^) ^ kT, or in volts 



