Respuesta :
Answer:
(a) 2.5 cm
(b) Yes
Solution:
As per the question:
Mass of Uranium-235 ion, m = [tex]3.95\times 10^{- 25}\ kg[/tex]
Mass of Uranium- 238, m' = [tex]3.90\times 10^{- 25}\ kg[/tex]
Velocity, v = [tex]3.00\times 10^{5}\ m/s[/tex]
Magnetic field, B = 0.250 T
q = 3e
Now,
To calculate the path separation while traversing a semi-circle:
[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]
The radius of the ion in a magnetic field is given by:
R = [tex]\frac{mv}{qB}[/tex]
[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]
[tex]\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})[/tex]
[tex]\Delta x = 2(\frac{m - m'}{qB}v)[/tex]
Now,
By putting suitable values in the above eqn:
[tex]\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[/tex]
[tex]\Delta x = 1.25\ cm[/tex]
(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.
