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A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from the velocity selector at a speed of 2.75×10⁵

m/s and enter a uniform magnetic field of 0.580 T directed perpendicularly to the velocity of the ions. What is the distance between the impact points formed on the photographic plate by singly charged ions of ²³⁵\
U and 238U ? cm

User Jahmar
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Final answer:

The distance between the impact points formed on the photographic plate by singly charged ions of uranium-235 and uranium-238 in a mass spectrometer is approximately 1.791 x 10^(-6) cm.

Step-by-step explanation:

The motion of charged particles in a magnetic field is described by the Lorentz force equation:

F=qvB

where:

F is the magnetic force,

q is the charge of the particle,

v is the velocity of the particle,

B is the magnetic field strength.

The magnetic force provides the centripetal force for the circular motion of the ions. The centripetal force (Fc) is given by:

Fc = mv2/r

where:

m is the mass of the particle,

v is the velocity of the particle,

r is the radius of the circular path.

Setting the magnetic force equal to the centripetal force, we get:

qvB= mv2/r

Solving for

r, the radius of the circular path, we find:

r= mv/qB

Now, the distance traveled by the ion on the photographic plate (d) is related to the circular path by the formula:

d=2πr

Substituting the expression for

r into the formula for d, we get:

d=2π mv/qB

Now, let's calculate the distance d for both 235U and 238U ions. The masses and charges for 235U and 238U are different, so we need to take those values into account.

For 235U:

Mass (m235):

235×1.6605×10−27kg (mass of a proton)

Charge (q235): 1×e (where e is the elementary charge, approximately 1.602×10−19C)

For 238U:

Mass (m238):

238×1.6605×10−27kg

Charge (q238): 1×e

Now, we can substitute these values into the formula for d:

d = 2π m235v/ q238B

d' = 2π m235v/ q238B

Now, plug in the given values:

v=2.75×105m/s

B=0.580 T

Calculate d and d′, and the difference between them will give you the distance between the impact points on the photographic plate. Note that the masses are very close, so the difference in distances should be small.

User Kenny Eliasson
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