The plates of a Thomson q / m apparatus are 6.0 cm long and are separated by 1.0

Question

The plates of a Thomson q/m apparatus are 6.0 cm long and are separated by 1.0 cm. The end of the plates is 30.0 cm from the tube screen. The kinetic energy of the electrons is 2.6 keV.

(a) If a potential of 25.0 V is applied across the deflection plates, by how much will the beam deflect?

_____________mm

(b) Find the magnitude of the crossed magnetic field that will allow the beam to pass through undeflected.

_____________T

i got part b right, but part a got it wrong.

and some helped me with it, and he ended by the answer below, but its wrong.

a) Since K = 2.6 keV = 2.6*1.60x10^-16J/keV = 4.16x10^-16 => v = sqrt(2*K/m)

= sqrt(2*4.16x10^-16/9.11x10^-31) = 30220557 m/s

The time between the plates is

t = l/v

= 0.060m/30220557

= 0.00000000198 s

The acceleration is

F/m

= E*q/m

= V*q/(d*m)

= 25*1.60x10^-19/(0.01*9.11x10^-31)

= 4.39x10^14 m/s^2

so y = 1/2*a*t^2

= 1/2*4.39x10^14*(0.00000000198)^2

= 0.000860 m

Now the particle leaves the plates with vx = 30220557 m/s
and vy = a*t = 4.39x10^14*0.00000000198 = 86920 m/s

so the angle is arctan( 86920/30220557) = 0.16

So the drop over the 30cm is y = 0.30*tan(0.16) = 0.00083 m

So overall the particle moves

0.000860 + 0.00083

= 0.00169 m

= 1.69 mm (This is the deflection at the screen)

Explanation / Answer

a) Since K = 2.6keV = 2.6*1.60x10^-16J/keV = 4.16x10^-16

Kinetic energy K = 1/2 mv2 = 2.6 KeV

v = sqrt(2*K/m)

= sqrt(2*4.16x10^-16/9.11x10^-31) = 3.02 x10^7m/s

The time between the plates is t = l/v = 0.060m/3.02 x10^7 = 0.1986x10^-7s = 1.98610-9 seconds

The acceleration is F/m = E*q/m = V*q/(d*m) = 25*1.60x10^-19/(0.01*9.11x10^-31) =4.39 x10^14m/s^2

so y = 1/2*a*t^2 = 1/2*2.439x10^14*(2.15x10^-9)^2 = 8.66x10^-4m (This is the deflections at the end of the plates)

Now the particle leaves the plates with vx = 3.02x10^7m/s
and vy = a*t = 4.39x10^14*1.986x10^-9 = 8.72x10^5m/s
so the angle is arctan( 8.72x10^5/3.02x10^7) = 1.650

So the drop over the 30cm is y = 0.30*tan(1.65) = 8.66 x10^-3m

So overall the particle moves 8.66x10^-4 + 86.6 x10^-4 = 9.526 x10^-3m (This is the deflection at the screen)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.