# 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?

Apr 2, 2016

$0.875 c m$

#### Explanation:

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

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

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

$r = \frac{m \times 1.05 \times {10}^{5}}{1.60 \times {10}^{-} 19 \times 0.75}$

$r = m \times 8.75 \times {10}^{23}$

1. For $\text{U} 235$:
${r}_{235} = 3.9 \times {10}^{-} 25 \times 8.75 \times {10}^{23}$
$= 0.34125 m$
2. For $\text{U} 238$:
${r}_{238} = 3.95 \times {10}^{-} 25 \times 8.75 \times {10}^{23}$
$= 0.345625 m$

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

$= 2 \times \left(0.345625 - 0.34125\right) = 0.00875 m$
$= 0.875 c m$