# 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 - 25 kg and charge 3.20 times 10 - 19 C from related species. The ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.00 m. After traveling through 180 degree and passing through a slit of width 1.00 mm and height 1.00 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 100 mg of material per hour, calculate (b) the current of the desired ions in the machine and (c) the thermal energy produced in the cup in 1.00 h.

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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 - 25 kg and charge 3.20 times 10 - 19 C from related species. The ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.00 m. After traveling through 180 degree and passing through a slit of width 1.00 mm and height 1.00 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 100 mg of material per hour, calculate (b) the current of the desired ions in the machine and (c) the thermal energy produced in the cup in 1.00 h.

2020-11-27
a) The magnitude of the manetic field is solvable using theequation $$\displaystyle{r}={m}\times\frac{{V}}{{{\left|{q}\right|}\times{B}}}$$ , where r is the radius of theion's motion, m is the ion's mass, V is the ion's velocity, q isthe ion's charge, and B is the magnetic field acting on theion.
$$\displaystyle{\frac{{{1}}}{{{2}}}}{M}{\left({V}{e}{l}{o}{c}{i}{t}{y}\right)}^{{{2}}}={q}{\left({V}{o}{<}{a}\ge\right)}$$</span>
PSK\frac{1}{2}\times (3.92\times 10^{-25}kg) \times V^{2} = (3.2 \times 10^{-19} C) \times 100kVZSK
$$\displaystyle{V}={4.04}\times{10}^{{{5}}}\frac{{m}}{{s}}$$
$$\displaystyle{r}={m}\times\frac{{V}}{{{\left|{q}\right|}\times{B}}}$$
$$\displaystyle{1}{m}={\frac{{{\left({3.92}\times{10}^{{-{25}}}{k}{g}\right)}\times{4.04}\times{10}^{{{5}}}\frac{{m}}{{s}}}}{{{\left({3.2}\times{10}^{{-{19}}}{C}\times{B}\right)}}}}$$
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
$$\displaystyle{\frac{{{\left\lbrace{.1}\times{10}^{{-{3}}}{k}{g}\right\rbrace}{\left\lbrace{3.92}\times{10}^{{-{25}}}{k}{g}\right\rbrace}}}{\times}}{\frac{{{\left\lbrace{3}\times{10}^{{-{19}}}{C}\right\rbrace}{\left\lbrace{3600}{s}\right\rbrace}}}{=}}{I}$$
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.
PSK\frac{1}{2}M(Velocity)^2 \times (massCollected / massPerIon)= EnergyDisspated \frac{1}{2} \times \frac{1}{2} \times (3.92\times 10^{-25}kg) \times (4.04\times 10^{5})^{2} \times \frac{.1\times 10^{-3}kg}{3.92\times 10^{-25}kg}) = EnergyZSK
Energy = $$\displaystyle{8.16}\times{10}^{{{9}}}{J}{o}\underline{{e}}{s}$$

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