Two isotoprs of uranium U235(3.9x10^-25 kg) , U238(3.95x10^-25kg) into spectrometer (1.05x10^5 m/s). Each isotope is singly ionized and B field(0.75T). What is the distance between the two isotopes?

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Two isotoprs of uranium U235(3.9x10^-25 kg) , U238(3.95x10^-25kg) into spectrometer (1.05x10^5 m/s). Each isotope is singly ionized and B field(0.75T). What is the distance between the two isotopes?

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Answer 1 · 2016-04-02T07:36:23

$0.875 c m$ $0.875cm$

Explanation:

Given that two isotopes of Uranium $U 235 (3.9 \times 10^{- 25} k g) and U 238 (3.95 \times 10^{- 25} k g)$ $"U"235(3.9xx10^-25 kg) and "U"238(3.95xx10^-25kg)$ into spectrometer.
Velocity of each singly ionized atom is $1.05 \times 10^{5} m / s$ $1.05xx10^5 m//s$
Force $\to F$ $vecF$ on a charged particle due to magnetic field $= q (\to v \times \to B)$ $=q(vecvxxvecB)$

Since the magnetic field is perpendicular to the direction of motion in a spectrometer
$∣ ∣ ∣ \to F ∣ ∣ ∣ = q . ∣ ∣ \to v ∣ ∣ . ∣ ∣ ∣ \to B ∣ ∣ ∣$ $|vecF|=q.|vecv|.|vecB|$ , singly ionized atom has charge $1.60 \times 10^{- 19} C$ $1.60xx 10^-19C$

This magnetic force produces circular motion of the ion.
Centripetal force $\frac{m . v^{2}}{r} = ∣ ∣ ∣ \to F ∣ ∣ ∣$ $(m.v^2)/r=|vecF|$ , where $r$ $r$ is the radius of circle in which ion moves.
$\frac{m . v^{2}}{r} = q . ∣ ∣ \to v ∣ ∣ . ∣ ∣ ∣ \to B ∣ ∣ ∣$ $(m.v^2)/r=q.|vecv|.|vecB|$ , Solving for $r$ $r$
$r = \frac{m . ∣ ∣ \to v ∣ ∣}{q . ∣ ∣ ∣ \to B ∣ ∣ ∣}$ $r=(m.|vecv|)/(q.|vecB|)$

$r = \frac{m \times 1.05 \times 10^{5}}{1.60 \times 10^{- 19} \times 0.75}$ $r=(mxx1.05xx10^5)/(1.60xx 10^-19xx0.75)$

$r = m \times 8.75 \times 10^{23}$ $r=mxx8.75xx10^23$

For $U 235$ $"U"235$ :
$r_{235} = 3.9 \times 10^{- 25} \times 8.75 \times 10^{23}$ $r_235=3.9xx10^-25xx8.75xx10^23$
$= 0.34125 m$ $=0.34125m$
For $U 238$ $"U"238$ :
$r_{238} = 3.95 \times 10^{- 25} \times 8.75 \times 10^{23}$ $r_238=3.95xx10^-25xx8.75xx10^23$
$= 0.345625 m$ $=0.345625m$

Distance between the two isotopes is equal to difference between the two diameters

![http://buphy.bu.edu](https://useruploads.socratic.org/Rb5WBK3TUigbNrCklF3B_12a.GIF)
$= 2 \times (0.345625 - 0.34125) = 0.00875 m$ $=2xx(0.345625-0.34125)=0.00875m$
$= 0.875 c m$ $=0.875cm$

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Two isotoprs of uranium U235(3.9x10^-25 kg) , U238(3.95x10^-25kg) into spectrometer (1.05x10^5 m/s). Each isotope is singly ionized and B field(0.75T). What is the distance between the two isotopes?

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Explanation:

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