Question

(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is used as reactor fuel.) The masses of the ions are 3.90x10—25 kg and 3.95x10—25 kg , respectively, and they travel at 3.00x105 m/s in a 0.250-T field. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238.

Answer 1

Answer:

(a) 2.5 cm

(b) Yes

Solution:

As per the question:

Mass of Uranium-235 ion, m = $3.95\times 10^{- 25}\ kg$

Mass of Uranium- 238, m' = $3.90\times 10^{- 25}\ kg$

Velocity, v = $3.00\times 10^{5}\ m/s$

Magnetic field, B = 0.250 T

q = 3e

Now,

To calculate the path separation while traversing a semi-circle:

$\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})$

The radius of the ion in a magnetic field is given by:

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

$\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})$

$\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})$

$\Delta x = 2(\frac{m - m'}{qB}v)$

Now,

By putting suitable values in the above eqn:

$\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm$

$\Delta x = 1.25\ cm$

(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.