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Sagot :
Answer:
a) [tex] v_{U-235} = 2.68 \cdot 10^{5} m/s [/tex]
[tex]v_{He-4} = -1.57 \cdot 10^{7} m/s[/tex]
b) [tex] E_{He-4} = 8.23 \cdot 10^{-13} J [/tex]
[tex] E_{U-235} = 1.41 \cdot 10^{-14} J [/tex]
Explanation:
Searching the missed information we have:
E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J
m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg
m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg
a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:
Linear momentum:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235} [/tex]
Since the plutonium nucleus is originally at rest, [tex]v_{Pu-239} = 0[/tex]:
[tex] 0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235} [/tex]
[tex] v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}} [/tex] (1)
Kinetic Energy:
[tex] E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2} [/tex]
[tex] 2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2} [/tex]
[tex] 1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2} [/tex] (2)
By entering equation (1) into (2) we have:
[tex] 1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2} [/tex]
[tex] 1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2} [/tex]
Solving the above equation for [tex]v_{U-235}[/tex] we have:
[tex] v_{U-235} = 2.68 \cdot 10^{5} m/s [/tex]
And by entering that value into equation (1):
[tex]v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s[/tex]
The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.
b) Now, the kinetic energy of each nucleus is:
For He-4:
[tex]E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J[/tex]
For U-235:
[tex] E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J [/tex]
I hope it helps you!
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