Question

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?

Answer 1

`0.875cm`

Explanation:

Given that two isotopes of Uranium `"U"235(3.9xx10^-25 kg) and "U"238(3.95xx10^-25kg)` into spectrometer.
Velocity of each singly ionized atom is `1.05xx10^5 m//s`
Force `vecF`on a charged particle due to magnetic field `=q(vecvxxvecB)`

Since the magnetic field is perpendicular to the direction of motion in a spectrometer
`|vecF|=q.|vecv|.|vecB|`, singly ionized atom has charge `1.60xx 10^-19C`

This magnetic force produces circular motion of the ion.
Centripetal force `(m.v^2)/r=|vecF|`, where `r` is the radius of circle in which ion moves.
`(m.v^2)/r=q.|vecv|.|vecB|`, Solving for `r`
`r=(m.|vecv|)/(q.|vecB|)`

`r=(mxx1.05xx10^5)/(1.60xx 10^-19xx0.75)`

`r=mxx8.75xx10^23`

1. For `"U"235`:
   `r_235=3.9xx10^-25xx8.75xx10^23`
   `=0.34125m`
2. For `"U"238`:
   `r_238=3.95xx10^-25xx8.75xx10^23`
   `=0.345625m`

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

`=2xx(0.345625-0.34125)=0.00875m`
`=0.875cm`