contestada

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus , the latter of which is also radioactive and will itself decay some time later. The energy emitted in the plutonium decay is and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is , while that of the uranium is (note that the ratio of the masses is 4 to 235).
(a) Calculate the velocities of the two nuclei, assuming the plutonium nucleus is originally at rest.
(b) How much kinetic energy does each nucleus carry away

Respuesta :

whitneytr12

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!                                                                                    

Otras preguntas