A proton [q,m], a deuteron[q,2m] and an alpha particle [2q,2m] walk into a bar..

Question

A proton [q,m], a deuteron[q,2m] and an alpha particle [2q,2m] walk into a bar...no wait..they are accelerated from rest by the same potential difference [V] and enter the same region of uniform magnetic field [B] moving perpendicular to the direction of B.

a) Calculate the ratio of the proton's kinetic energy to that of the alpha particle.

b) Calculate the ratio of the deuteron's kinetic energy to that of the alpha particle.

c) If the proton traces a circle of radius 10 cm what is the radius of the deuteron's path?

d) If the proton traces a circle of radius 10 cm what is the radius of the alpha particle's path?

Explanation / Answer

Part A)

From conservation of energy PE = KE

PE = qV so for the proton to the alpha, you get

qV/2qV

The q and V cancel, so the ratio is 1/2

Part B)

The deutron is the same since the charges are q to 2q

So that ratio is also 1/2

Part C)

The radius is found from the formula r = mv/qB

The velocity comes from the KE, so

qV = .5mv2

v = sqrt(2qV/m)

So the deutron velocity to the proton velocity is

sqrt (2qV/2m)/sqrt(2qV/m)

That simplifies to sqrt (1/2)

The r to r ratio for deutron to proton is

(2m/qB)/(m/qB) times sqrt(1/2)

That is 2sqrt(1/2)

That is 1.414

Then take 1.414 times 10 and get 14.14 cm

Part D)

Same idea for the alpha particle

The radius is found from the formula r = mv/qB

The velocity comes from the KE, so

qV = .5mv2

v = sqrt(2qV/m)

So the alpha velocity to the proton velocity is

sqrt (4qV/2m)/sqrt(2qV/m)

That simplifies to 1

The r to r ratio for alpha to proton is

(2m/2qB)/(m/qB) times 1

That is 1

Then take 1 times 10 and get 10 cm

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