How much of the atom's internal energy is released?

In summary: The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units=> Thorium-234 has a mass of 234 amu, and α-4 has a mass of 4 amu.You also know 1 amu = 1.66 × 10−27 kgHow can you combine those two things to get the masses of the nuclei in kg?To find the masses in kilograms, we can use the conversion factor of 1 amu = 1.66 × 10−27 kg. For thorium-234, we can calculate its mass in kilograms by multiplying 234
  • #1
emily081715
208
4

Homework Statement


A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, α-4. The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units (1 amu = 1.66 × 10−27 kg). When an uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.9 × 105 m/s.How much of the uranium atom's internal energy is released in the breakup?

Homework Equations


k=1/2mv^2

The Attempt at a Solution


i assumed that the energy released would come form the amount of kinetic energy of the thorium - the kinetic energy of the alpha particle. this got me and answer of -1.60x 10^-14. i believe that i hit the problem when calculating the mass to use of the atoms except i am unsure if this is even the correct way to approach the question
 
Physics news on Phys.org
  • #2
emily081715 said:
i assumed that the energy released would come form the amount of kinetic energy of the thorium - the kinetic energy of the alpha particle.
Both the kinetic energies come from the energy released in the decay. You have to sum the energies.
How did you get the kinetic energy of the alpha particle?
emily081715 said:
i believe that i hit the problem when calculating the mass to use of the atoms
What did you use and where do you expect a problem?

Your answer is missing units.
 
  • #3
mfb said:
Both the kinetic energies come from the energy released in the decay. You have to sum the energies.
How did you get the kinetic energy of the alpha particle?
What did you use and where do you expect a problem?

Your answer is missing units.
i looked up the mass of an alpha particle and it was 6.64424. 10-27 kg. The i found the kinetic energy of that and it was 2.79390292x10^-16. next i found the mass of thorium by multipying 234 by 1.66 × 10−27 kg. Then i found the kinetic energy of thorium and add it to the alpha particle kinetic energy. the answer was 6.86x10^10 J. this is inncorrect. can you tell me where my error is
 
  • #4
emily081715 said:
i looked up the mass of an alpha particle and it was 6.64424. 10-27 kg. The i found the kinetic energy of that and it was 2.79390292x10^-16. next i found the mass of thorium by multipying 234 by 1.66 × 10−27 kg. Then i found the kinetic energy of thorium and add it to the alpha particle kinetic energy. the answer was 6.86x10^10 J. this is inncorrect. can you tell me where my error is
Please post all your working.
 
  • #5
haruspex said:
Please post all your working.
ALPHA PARTICLE : mass= 6.64424. 10-27 kg(1.66 × 10−27 kg)=1.1029x10^-53 amu
k=(0.5)(1.1029x10^-53)(-2.9 × 10^5)^2=4.63x10^-43 J
THORIUM: 234(1.66 x10^-27)=3.88x10^-25 amu
k=(0.5)(3.88x10^-25)(-2.9 × 10^5)^2=1.6315x10^-14 J
1.6315x10^-14 J+4.63x10^-43 J=1.63x10^-15 J
 
  • #6
emily081715 said:
1.1029x10^-53 amu
That is not right.
emily081715 said:
k=(0.5)(1.1029x10^-53)(-2.9 × 10^5)^2=4.63x10^-43 J
You have to calculate the speed of the alpha particle first. It does not have the same speed as the thorium nucleus.
emily081715 said:
234(1.66 x10^-27)=3.88x10^-25 amu
That is wrong as well because you keep mixing units randomly.
 
  • #7
mfb said:
That is not right.You have to calculate the speed of the alpha particle first. It does not have the same speed as the thorium nucleus.
That is wrong as well because you keep mixing units randomly.
can you please explain more where i made my error instead of just saying its wrong
 
  • #8
emily081715 said:
can you please explain more where i made my error instead of just saying its wrong
is the mass of the thorium atom 234.0436 amu?
 
  • #9
emily081715 said:
is the mass of the thorium atom 234.0436 amu?
Hey Emily! ;)

Yep. So where you had amu before, it should be kg. Oh, and an alpha particle is 4.00 amu (2 protons and 2 neutrons).

We can find the speed of the alpha particle from the total momentum. It must be zero since the Uranium atom was at rest.
 
  • Like
Likes Greg Bernhardt
  • #10
I like Serena said:
Hey Emily! ;)

Yep. So where you had amu before, it should be kg. Oh, and an alpha particle is 4.00 amu (2 protons and 2 neutrons).

We can find the speed of the alpha particle from the total momentum. It must be zero since the Uranium atom was at rest.
so am i able to use amu for mass?
 
  • #11
emily081715 said:
so am i able to use amu for mass?

Yep. (Nod)
It's a unit just like grams - just smaller, and tuned to atomic masses.
 
  • #12
I like Serena said:
Yep. (Nod)
It's a unit just like grams - just smaller, and tuned to atomic masses.
would that make the kinetic energy of the thorium atom : k=(0.5)(234)(-2.9 × 10^5)^2=9.83x10^12J?
 
  • #13
emily081715 said:
would that make the kinetic energy of the thorium atom : k=(0.5)(234)(-2.9 × 10^5)^2=9.83x10^12J?

That is a LOT of joules for such a microscopic particle. :oldsurprised:
To get joules we have to convert the amu unit to kg. That is, multiply 234 amu by 1.66x10^-27 kg/amu.
 
  • #14
I like Serena said:
That is a LOT of joules for such a microscopic particle. :oldsurprised:
To get joules we have to convert the amu unit to kg. That is, multiply 234 amu by 1.66x10^-27 kg/amu.
"THORIUM: 234(1.66 x10^-27)=3.88x10^-25 amu
k=(0.5)(3.88x10^-25)(-2.9 × 10^5)^2=1.6315x10^-14 J"
so this process from earlier was correct for thorium?
 
  • #15
emily081715 said:
"THORIUM: 234(1.66 x10^-27)=3.88x10^-25 amu
k=(0.5)(3.88x10^-25)(-2.9 × 10^5)^2=1.6315x10^-14 J"
so this process from earlier was correct for thorium?
The process was correct. The unit was wrong.
It's 234 amu = 3.88×10-25 kg.
 
Last edited:
  • #16
I like Serena said:
The process was correct. The unit was wrong.
It's 234 amu = 3.88x10^-25 kg.
but the amount of kinetic energy is still right?
 
  • #17
emily081715 said:
can you please explain more where i made my error instead of just saying its wrong
I said "use units" twice already. And not in the way of "I'll add some unit to the final result and hope that it is right." Use them consistently and you'll spot the majority of your errors on your own.

Let's start from scratch with the masses:
emily081715 said:
The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units
=> Thorium-234 has a mass of 234 amu, and α-4 has a mass of 4 amu.
You also know 1 amu = 1.66 × 10−27 kg

How can you combine those two things to get the masses of the nuclei in kg?
emily081715 said:
but the amount of kinetic energy is still right?
The kinetic energy for thorium is right, the one for the alpha particle is not.
 
  • #18
mfb said:
I said "use units" twice already. And not in the way of "I'll add some unit to the final result and hope that it is right." Use them consistently and you'll spot the majority of your errors on your own.

Let's start from scratch with the masses:
=> Thorium-234 has a mass of 234 amu, and α-4 has a mass of 4 amu.
You also know 1 amu = 1.66 × 10−27 kg

How can you combine those two things to get the masses of the nuclei in kg?
The kinetic energy for thorium is right, the one for the alpha particle is not.
okay so the mass of the alpha particle is 4amu(1.66x10^-27kg)= 6.64x 10^-27 kg
 
  • #19
emily081715 said:
okay so the mass of the alpha particle is 4amu(1.66x10^-27kg)= 6.64x 10^-27 kg
now i need to solve for speed of the alpha particle. can i do that by using mommetum ?
 
  • #20
emily081715 said:
now i need to solve for speed of the alpha particle. can i do that by using mommetum ?
so i used the equation 1/2mv^2=1/2mv^2
√(1.61315x10^-14)/(0.5)(6.64x10^-27)
v=2.2x10^6 m/s

K=(0.5)(6.64x 10^-27)(2.2x10^6)^2
=1.60x10^-14J

sum: 1.60x10^-14J+1.63x10^-15 J
=1.76x10^-14J

Is this correct?
 
  • #21
emily081715 said:
okay so the mass of the alpha particle is 4amu(1.66x10^-27kg)= 6.64x 10^-27 kg

Properly that should be:

4amu(1.66x10^-27kg/amu)= 6.64x 10^-27 kg

or:

4amu=4(1.66x10^-27kg)= 6.64x 10^-27 kg

emily081715 said:
now i need to solve for speed of the alpha particle. can i do that by using mommetum ?
Yes.

emily081715 said:
so i used the equation 1/2mv^2=1/2mv^2
Is this correct?

Nope. Conservation of momentum means ##m_{Th} v_{Th} + m_\alpha v_\alpha = 0##.
 
  • #22
okay using that equation i got the speed of an alpha particle as 1.6945x107 m/s.
then i plugged the into the kinetic energy formula and found that the kinetic energy of the particle was 1.63x10-15 J.
adding this to the energy of thorium, i got a final answer of 1.63x10-15 J. Is this correct?
 
  • #23
emily081715 said:
okay using that equation i got the speed of an alpha particle as 1.6945x107 m/s.
then i plugged the into the kinetic energy formula and found that the kinetic energy of the particle was 1.63x10-15 J.
adding this to the energy of thorium, i got a final answer of 1.63x10-15 J. Is this correct?

I get the same speed for the alpha particle, but a different kinetic energy.
Btw, if we add the kinetic energy of the Thorium particle to it, shouldn't we get a different number?
 

Related to How much of the atom's internal energy is released?

1. How is the internal energy of an atom released?

The internal energy of an atom is released through a process called nuclear fission, where the nucleus of the atom splits into two or more smaller nuclei, releasing a large amount of energy.

2. What factors affect the amount of energy released by an atom?

The amount of energy released by an atom depends on the type and structure of the atom's nucleus, as well as the type and amount of energy used to trigger the release.

3. How is the energy released by an atom measured?

The energy released by an atom is measured in units of joules (J) or electron volts (eV). These units represent the amount of energy needed to move one coulomb of charge through one volt of potential difference.

4. Can all atoms release energy?

No, not all atoms are capable of releasing energy. Only atoms with unstable nuclei, such as uranium and plutonium, are able to undergo nuclear fission and release energy.

5. What are the potential uses for the energy released by atoms?

The energy released by atoms has a variety of potential uses, including electricity generation, nuclear weapons, and nuclear medicine. It can also be used for scientific research and space exploration.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
2K
Replies
13
Views
487
  • Introductory Physics Homework Help
Replies
8
Views
7K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
23
Views
2K
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
885
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Back
Top