Respuesta :
Answer:
d = 0.030 m
Explanation:
The radius of curvature (r) of the two isotopes can be calculated using the following equation:
[tex] r = \frac{m*v}{q*B} [/tex]
Where:
- m: is the mass of the isotope, m₁ (C-12) = 1.993x10⁻²⁶ kg, m₂(C-13) = 2.159x10⁻²⁶ kg
- v: is the speed of the isotpe, v₁ (C-12) = v₂ (C-13) = 7.13x10⁵ m/s
- B: is the magnetic field = 0.5100 T
- q: is the charge of the isotopes, q₁ = q₂ = +1.6x10⁻¹⁹ C
The radius of the isotope carbon-12 is:
[tex] r_{1} = \frac{1.993 \cdto 10^{-26} kg*7.13 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.5100 T} = 0.174 m [/tex]
Simimlarly, the radius of the isotope carbon-13 is:
[tex] r_{2} = \frac{2.159 \cdto 10^{-26} kg*7.13 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.5100 T} = 0.189 m [/tex]
Finally, the spatial separation between the two isotopes after they have traveled through a half-circle ([tex]d_{T}[/tex]) is:
[tex] d_{T} = d_{2} - d_{1} = 2r_{2} - 2r_{1} = 2 (0.189 m - 0.174 m) = 0.030 m [/tex]
I hope it helps you!
The spatial separation between the two isotopes after they have traveled through a half-circle should be 0.030m.
Calculation of the spatial separation:
Since
r = m*v/q * B
here,
m is the mass of the isotope,
m₁ (C-12) = 1.993x10⁻²⁶ kg,
m₂(C-13) = 2.159x10⁻²⁶ kg
v is the speed of the isotpe,
v₁ (C-12)
v₂ (C-13) = 7.13x10⁵ m/s
B is the magnetic field = 0.5100 T
q is the charge of the isotopes, q₁ = q₂ = +1.6x10⁻¹⁹ C
Now the radius for carbon 12 is
= 1.99310*^-26 * 7.13*10^15/1.6*10^19 * 0.5100 T
= 0.174m
And, the radius for carbon 13 is
= 2.15910^-26 * 7.13.10^5/1.6*10^-19*0.5100T
= 0.189m
So, here the spatial separation should be
= d_2-d_1
= 2r_2 - 2r_1
= 2(0.189m - 0.174m)
= 0.030m
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