Question #87826
THe plates of a thompson q/m apparatus are 5 cm long and separated by 1 cm. The end of the plates are 25 cm from tube screen. The kinetic energy of the electrons is 2000 eV.
a) If a potential of 20 V is applied across the deflection plates, by how much will the beam deflect
1
Expert's answer
2019-04-10T09:52:06-0400

Electron mass: m = 511 keV = 511000 eV

Speed of the electron:

E=mv2/2v=2EmE=mv^2/2 \Rightarrow v = \sqrt{\frac{2E}{m}}

Amount of time, when electron was in apparatus:t=a/v,a=5cm=0.05mt =a/v,\, a =5\text{cm} = 0.05 \text{m}

Electric force, applied to electron:

F=eU/d,U=20V,d=1textcm=0.01mF = e U / d,\, U = 20\, \text{V}, \, d = 1\, text{cm} = 0.01\, \text{m}

where e - elementary charge.

From second Newtons law follows that change of momentum equals to


Δp=mΔv=Ft=eUavdΔv=eUavdm\Delta p = m \Delta v = F \cdot t = \frac{eUa}{vd} \Rightarrow \Delta v = \frac{eUa}{vdm}

Change of speed is comperatively small:


ϕ=Δv/v=eUav2dm=eUa2Ed=200.0540000.01=0.025\phi = \Delta v / v = \frac{eUa}{v^2 d m} =\frac{eUa}{2 E d} = \frac{20\cdot 0.05}{4000 \cdot 0.01} = 0.025

Δv\Delta v perpendicular to the initial direction of speed. Therefore, direction of velocity would be changed on angle ϕ\phi

Beam would be deflected by distance:


l=ϕL=eUa2EdL=0.02525=0.625cm=6.25mml = \phi \cdot L = \frac{eUa}{2 E d} L = 0.025 \cdot 25 = 0.625 \, \text{cm} = 6.25 \, \text{mm}


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