I Help Chap & Begin Date: 3/26/2017 12:00:00 AM-Due Date: 10/29/2017 11:59:00 PM

Question

I Help Chap & Begin Date: 3/26/2017 12:00:00 AM-Due Date: 10/29/2017 11:59:00 PM End Date: 12/15/2017 12:00:00 AM (11%) Problem 6: One of the waste products of a nuclear reactor is plutonium-239(239Pu). This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus (He +23SU), the latter of which is also radioactive and will itself decay some time later. The energy emitted in the plutonium decay is 8.40 x 10 13 J and is entirely converted into the kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus s 66s-1027 kg while that of the uraniumis 392 kg (note that the ratio of the masses is 4 to 235). ally at rest. He Potential 96 cotan)asin acos) (4% per attempt) detailed view atan) acotan sinh0 cosh0 tanh) cotanh) I give upt Hints: 4% deduction per hint. Hits on per 25% Part (b) Calculate the speed of the U, in meters per second assuming the plutonium nucleus is originally at rest. 25% Part (c) How much kinetic energy, in Joules, does the He nucleus carry away? 25% Part (d) How much kinetic energy, in joules, does the U nucleus carry away?

Explanation / Answer

The large nucleus splits into 2 smaller ones:

helium: mass=m velocity=v

uranium: mass=m velocity=v

From conservation of momentum:

final momentum = initial momentum

mv + mv = 0

v = -(m/m)v = -(3.92×10² /6.68x10²)v = -58.68v

From conservation of energy:

½mv² + ½mv² = 8.40x10¹³

For the rest of the arithmetic I'll use exponential notation, e.g. 8.40x10¹³ is written 8.40E-13. This is easier to type and looks neater.

Substitute for m, m and v:

½* 6.68E-27*(-58.68v)² + ½(3.92E-25)v² = 8.40E-13

1.150E-23v² + 1.96E-25v² = 8.40E-13

1.1696E-23v² = 8.40E-13

v² = 7.1819E10

(b) v = 2.68E5 = 2.68x10 m/s

That’s the uranium nucleus’s speed.

(a) v= -58.68v = -58.68*2.68E5 = 1.57E7 = 1.57x10 m/s

That’s the helium nucleus’s speed

(c) Kinetic energy carried away be helium nucleus is:

½mv² = ½* 6.68E-27*(1.57E7)² = 8.23E-13 = 8.23x10¹³ J

(d) Kinetic energy carried away be uranium nucleus is:

½mv² = ½*(3.92E-25)*(2.68E5)² = 1.41E-14 = 1.41x10¹J

[Note that these 2 energies add up to

8.23E-13 + 1.41E-14 = 8.37E-14J

They should add to give 8.40E-14J, so there is a small error, probably rounding error, in the arithmetic. If you work through it for yourself to more decimal places for the intermediate steps, you should be able to remove the error.]

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