Respuesta :
Answer:
The path separation is 0.089 m
Solution:
As per the question:
Mass of [tex]O_{16}[/tex], m = 2.66\times 10^{- 26}\ kg[/tex]
Ratio of the masses of [tex]O_{16}[/tex] to [tex]O_{18}[/tex], R = 16:18
Velocity of the masses, v = [tex]5.15\times 10^{6}\ m/s[/tex]
Magnetic field, B = 1.2 T
Now,
The separation between the path after completing a semicircle can be calculated as:
Mass of [tex]O_{18}[/tex], m' = mR = [tex]2.66\times 10^{- 26}\times \frac{18}{16} = 2.99\times 10^{- 26}\ kg[/tex]
Here, magnetic force provides the necessary centripetal force to traverse the semi-circle:
[tex]F_{B} = F_{c}[/tex]
[tex]qvB = \frac{mv^{2}}{r}[/tex]
where
[tex]F_{c} = \frac{mv^{2}}{r}[/tex]
[tex]F_{B} = qvB[/tex] = Magnetic force
q = e = electronic charge = [tex]1.6\times 10^{- 19}\ C[/tex]
v = Velocity
B = Magnetic field
r = Radius
Now, the radius from the above eqn comes out to be:
[tex]r = \frac{mv}{qB}[/tex]
Now, for [tex]O_{16}[tex]:
[tex]r_{16} = \frac{mv}{eB}[/tex]
[tex]r_{16} = \frac{2.66\times 10^{- 26}\times 5.15\times 10^{6}}{1.6\times 10^{- 19}\times 1.2} = 0.713\ m[/tex]
Now, for [tex]O_{18}[tex]:
[tex]r_{16} = \frac{m'v}{eB}[/tex]
[tex]r_{16} = \frac{2.99\times 10^{- 26}\times 5.15\times 10^{6}}{1.6\times 10^{- 19}\times 1.2} = 0.802\ m[/tex]
Now, the path separation is given by:
[tex]d = r_{18} - r_{16} = 0.802 - 0.713 = 0.089\ m[/tex]
