In Stern's experiment (1920) silver atoms emitted by a heated filament passed through a slit and were deposited on the cooled wall of an outer cylinder (Fig.). When the system was rotated at high speed there was a deflection of the slit's image. The apparatus was first rotated in one direction and then in the opposite direction, and the distance between the deflected images was measured. Find this distance if the radius of the internal cylinder is 2.0 cm and of the external one 8.0 cm. The speed of rotation is 2700 r.p.m. and the filament temperature `960^@C`.
 
Estimate the errors of measurement, if the width of the slit is 0.5 mm.

Answer

5.1 mm ; 40%

Explanation

The total displacement of the image is `l = 2r_2omegat`, where t is the time of transit of the molecule between the cylinders, i.e. `t = = (r_2 - r_1) sqrt(v)` . The root-mean-square velocity of the silver atoms is
`sqrt((3RT)/(M)) = sqrt((3 xx 8.3 xx 10^3 xx 1233)/(108)) = 532 m//s`
Hence `l = 2 omega r_2 (r_2 - r_1) sqrt(v)`
 To assess the error, calculate the width of the undisplaced image of the slit:
`a/b = (r_2)/(r_1),` whence `a = (l.r_2)/(r_1) = (0.5 xx 8)/(2) = 2 mm`
The broadening of the displaced images is still greater because of the Maxwell distribution of the molecular velocities. Hence the error in measurement is `Delta ge 2a//2 = a`, the relative error being
`epsilon = - Delta/l ge a/l = (2 xx 100%)/(5) = 40%`
This shows that the Stern experiment could produce only qualitative data on the molecular velocities.