Transcripted image text displayed Uranium ions with a mass of 3.92 x 10–25 kg and a charge of 3.20 x 10–19 C are separated from similar species using a specific commercial mass spectrometer (see Sample Problem 28–3). The ions enter a uniform magnetic field after being propelled through a potential difference of 100 kV, where they are bent along a path with a radius of 1 m. They are gathered in a cup after turning 180 degrees and going through a slot that is 1 mm wide and 1 cm high. (A) What is the strength of the separator's (perpendicular) magnetic field? Calculate (b) the current of the required ions in the machine and (c) the thermal energy produced in the cup in 1.00 hours if the machine is used to separate out 100 mg of material each hour.

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Transcripted image text displayed Uranium ions with a mass of 3.92 x 10–25 kg and a charge of 3.20 x 10–19 C are separated from similar species using a specific commercial mass spectrometer (see Sample Problem 28–3). The ions enter a uniform magnetic field after being propelled through a potential difference of 100 kV, where they are bent along a path with a radius of 1 m. They are gathered in a cup after turning 180 degrees and going through a slot that is 1 mm wide and 1 cm high. (A) What is the strength of the separator&#039;s (perpendicular) magnetic field? Calculate (b) the current of the required ions in the machine and (c) the thermal energy produced in the cup in 1.00 hours if the machine is used to separate out 100 mg of material each hour.

Jozlyn · Accepted Answer

a) The magnitude of the manetic field is solvable using theequation r=m×V|q|×B , where r is the radius of theion&#039;s motion, m is the ion&#039;s mass, V is the ion&#039;s velocity, q isthe ion&#039;s charge, and B is the magnetic field acting on theion. 12M(Velocity)2=q(Vo&amp;lt;a≥) 12×(3.92×10−25kg)×V2=(3.2×10−19C)×100kV V=4.04×105ms r=m×V|q|×B 1m=(3.92×10−25kg)×4.04×105ms(3.2×10−19C×B) B = 2.02Tesla b) Since current is measured in coulombs per second, the totalcharge moved per hour is the mass per hour over the mass per ion(which gives us the number of ions per hour), multiplied by thecharge per ion. This value is then divided by 3600seconds (perhour) to attain the current. (massCollected / massPerIon) * CoulombPerIon / TimePeriod= Current {.1×10−3kg}{3.92×10−25kg}×{3×10−19C}{3600s}=I I = .0227Amps c) Energy dissipated in the cup is just a function of the kineticenergy dissipated for each ion times the number of ions collectedduring the period. 12M(Velocity)2×(mass Collected/massPerIon)=Energy Disspated12×12×(3.92×10−25kg)×(4.04×105)2×.1×10−3kg3.92×10−25kg)=Energy Energy = 8.16×109Joe―s

Show transcribed image text A certain commercial mass spectrometer (see Sample Problem 28 - 3) is used to separate uranium ions of mass 3.92 times 10

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