Question

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial ice. (The relative abundance of these oxygen isotopes is related to climatic temperature at the time the ice was deposited.) The ratio of the masses of these two ions is 16 to 18, the mass of oxygen-16 is 2.66 × 10-26 kg, and they are both singly charged and travel at 4.9 × 106 m/s in a 1.05 T magnetic field.

Answer 1

Answer:

The path separation is 0.10 m

Solution:

As per the question:

Mass of $O_{16}$, m = $2.66\times 10^{- 26}\ kg$

Ratio of the masses, R = 16:18

Velocity, v = $4.9\times 10^{6}\ m/s$

Magnetic field, B = 1.05 T

Now,

Mass of $O_{18}$, m' = m\frac{1}{R} = 2.66\times 10^{- 26}\times \frac{18}{16} = 2.99\times 10^{- 26}\ kg[/tex]

Now, centripetal force here is provided by the magnetic force on the charge 'q' and is given by:

$\frac{mv^{2}}{r} = qvB$

$r = \frac{mv}{qB}$

Now, for $O_{16}$:

$r = \frac{2.66\times 10^{- 26}\times 4.9\times 10^{6}}{1.6\times 10^{- 19}\times 1.05} = 0.77\ m$

Now, for $O_{18}$:

$r' = \frac{m'v}{qB}$

$r' = \frac{2.99\times 10^{- 26}\times 4.9\times 10^{6}}{1.6\times 10^{- 19}\times 1.05} = 0.87\ m$

Now, the separation distance of the path is given by:

$\Delat D = r' - r = 0.87 - 0.77 = 0.10\ m$