The plates of a thomson q/m apparatus are 6 cm long and are separated by 1.2cm.

Question

The plates of a thomson q/m apparatus are 6 cm long and are separated by 1.2cm. The end of the plates is 30 cm from the tube screen. The kinetic energy of the electrons is 2.80 keV. If a potential difference of 25.0 V is applied across the deflection plates, by how much will the point where the beam of electons strike the screen displaced? Please explain mathematics and theory. Explain what a Thomson q/m apparatus is, how it looks and what it does. Is it that thing with a cork that pops out when water boils, or something else? What s the structure of it. Please explain variables and equations.

Explanation / Answer

given x1 =6cm =0.06m

x2 =30cm =0.3m

KE =2.8KeV=2.8*103*1.6*10-19

V=25V

charge of electron q=1.6*10-19C

mass of electron me=9.11*10-31kg

velocity v=(2KE/me) =(2*2.8*103*1.6*10-19/(9.11*10-31) =3.14*107m/s

E=V/d =25/1.2*10-2 =2083.3V/m

t=x1/v =0.06/3.14*107=1.91*10-9s

y =(qE/m)[(1/2)(x1/v)2 +(x1x2/v2)]

y =(1.6*10-19*2083.3/9.11*10-31)[(1.91*10-9)2+(0.06*0.3/(3.14*107)2)]

y =7.35*10-3m or 7.35mm

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