A singly ionized Lithium atom and a doubly ionized Oxygen atom are each accelera

Question

A singly ionized Lithium atom and a doubly ionized Oxygen atom are each accelerated from rest through a potential difference of 50 MV and enters a cloud chamber where a magnetic field of 8.5T is directed perpendicular to the plane of their motion. As seen from above (Looking In the -z direction), the particles each curve to the Left. a) If the atoms are moving In the x direction when they enter the magnetic field, what is the direction of the field? b) What Is the ratio of the radius of curvature of the Lithium atom to that of the Oxygen atom? C) If each atom loses 1 pJ of energy for each cm traveled, how far does each atom travel in the cloud chamber before hatting? d) Calculate the average acceleration experienced by each particle as it come to rest, assuming it to be constant.

Explanation / Answer

a) from looking -z dirction particle curves to left that means

so at initial point force is in +y direction.

F = q ( v X B)

j = + (i x B)

B = -k

magnetic force is towrads -z direction.

b) qV = mv^2 /2

v = sqrt(2qV / m)

|F| = qvB

F = mv^2 / r

r = mv^2 / qvB = mv /qB

r = m sqrt(2qV / m) / qB

r = sqrt( 2V /qm^3) / B

r_lithum / r_oxy = sqrt(1/q_L *m_L^3) / sqrt(1/q_O *m_O^3)

= sqrt(q_O*m_O^3 / q_L*m_L^3)

qO = +2e
qL = +e
mL = 7m
mO = 16m

ration = sqrt(2 x 16^2 / 1 x 7^2) = 10.45

C) For lithium :

INitial Energy = qV = 1.6x10^-19 x 50 x 10^6 = 8 x 10^-12 J

energy lost for each 1 cm = 1 x 10-12 J

so it will travel = 8 cm

for Oxygen :

Initial energy = 2eV = 16 x 10^-12 J

d = 16 cm

d) For lithium :

mv^2 /2 = 8 x 10^-12

(7 x 1.673 x 10^-27) v^2 / 2 = 8 x 10^-12

v^2 = 1.366 x 10^15

a =v^2 / 2d = 1.366 x 10^15 / 2*0.08 = 8.54 x 10^15 m/s2

For oxygen:

mv^2 /2 = 8 x 10^-12

(16 x 1.673 x 10^-27) v^2 / 2 = 16 x 10^-12

v^2 = 1.195 x 10^15

a =v^2 / 2d = 1.195 x 10^15 / 2*0.16 = 3.74 x 10^15 m/s2

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