Energy released through alpha decay

In summary: Yes, I got it. After 1 half life, radon becomes Po + He, but after 2 half lives, radon becomes 2 Po + 2 He, right?In summary, radon-222 decays spontaneously by emitting an alpha particle, resulting in a daughter nucleus of polonium. The atomic masses of these isotopes are Rn-222: 222.0157 u, Po: 218.00896 u, and He: 4.00260 u. In a sample containing 3.0 x 107 nuclei of radon-222, after three half lives, 7/8 of the original nuclei will have decayed, resulting in 3/8 x 107
  • #1
songoku
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Homework Statement


The radioactive isotope radon-222 may decay spontaneously by emitting an alpha particle. The daughter nucleus is an isotope of polonium. The atomic masses of these isotopes are:
Rn-222 : 222.0157 u
Po : 218.00896 u
He : 4.00260 u

A sample of radon-222 contains 3.0 x 107 nuclei. The total energy that will be release in a time interval of 3 T1/2 through alpha decay of the radon-222 is:
a. 3.856 MeV
b. 5.785 x 107 MeV
c. 8.677 x 107 MeV
d. 1.012 x 108 MeV
e. 1.085 x 108 MeV

Homework Equations


E = mc^2
T1/2 = (ln 2) / λ

The Attempt at a Solution


If the question asks only for one half-life then the energy released = (mass of Rn - mass of Po - mass of He) x 931.5 MeV. Is this correct?

I don't know how the time interval of 3 half-life and number of nuclei play part in this question. Are they used to find energy released or they are only redundant information?

Thanks
 
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  • #2
songoku said:

Homework Statement


The radioactive isotope radon-222 may decay spontaneously by emitting an alpha particle. The daughter nucleus is an isotope of polonium. The atomic masses of these isotopes are:
Rn-222 : 222.0157 u
Po : 218.00896 u
He : 4.00260 u

A sample of radon-222 contains 3.0 x 107 nuclei. The total energy that will be release in a time interval of 3 T1/2 through alpha decay of the radon-222 is:
a. 3.856 MeV
b. 5.785 x 107 MeV
c. 8.677 x 107 MeV
d. 1.012 x 108 MeV
e. 1.085 x 108 MeV

Homework Equations


E = mc^2
T1/2 = (ln 2) / λ

The Attempt at a Solution


If the question asks only for one half-life then the energy released = (mass of Rn - mass of Po - mass of He) x 931.5 MeV. Is this correct?

I don't know how the time interval of 3 half-life and number of nuclei play part in this question. Are they used to find energy released or they are only redundant information?

Thanks
What happens during a half life?
 
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  • #3
PeroK said:
What happens during a half life?
Radon decays so the number of nuclei becomes half of original. The mass will also becomes 111.00785 u? The mass of He will stay the same as before?
How about mass of Po?

Thanks
 
  • #4
songoku said:
Radon decays so the number of nuclei becomes half of original. The mass will also becomes 111.00785 u? The mass of He will stay the same as before?
How about mass of Po?

Thanks

In one half life half of the original nuclei decay. So, what about after three half lives.

The atomic masses of the isotopes don't change!
 
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  • #5
PeroK said:
In one half life half of the original nuclei decay. So, what about after three half lives.
The nuclei of radon will be 3/8 × 107

The atomic masses of the isotopes don't change!
You mean the mass of radon will still be 222.0157 u? Does decay also change the mass of the nuclei?

Thanks
 
  • #6
songoku said:
The nuclei of radon will be 3/8 × 107You mean the mass of radon will still be 222.0157 u? Does decay also change the mass of the nuclei?

Thanks

I would have said: after three half lives 7/8 of the original nuclei will have decayed.

The Radon nuclei are decaying into different nuclei, so the original nuclei are either gone or still the same.

That doesn't seem like a particularly difficult concept!
 
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  • #7
PeroK said:
I would have said: after three half lives 7/8 of the original nuclei will have decayed.

So the number of nuclei, mass, activity will be 1/8 of original?

The Radon nuclei are decaying into different nuclei, so the original nuclei are either gone or still the same.
One simple basic question:
radon decays into Po by emitting alpha. Let say the number of nuclei of radon is x initially. After 1 half life, the number of nuclei of radon is 1/2 x and Po is 1/2 x. After 2 half life, radon is 1/4 x and Po is 3/4 x. Is this correct?
The particle "lost" by radon becomes Po? Alpha particle not taking some of particle "lost" by radon?

Thanks
 
  • #8
songoku said:
After 2 half life, radon is 1/4 x and Po is 3/4 x. Is this correct?
Yes.
songoku said:
The particle "lost" by radon becomes Po? Alpha particle not taking some of particle "lost" by radon?
I do not understand your terminology. The number of particles of radon lost equals the number of particles of Po created and equals the number of He particles created (ignoring the decay of Po).
 
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  • #9
haruspex said:
I do not understand your terminology. The number of particles of radon lost equals the number of particles of Po created and equals the number of He particles created (ignoring the decay of Po).
I am not sure I get what you mean. If the particles lost by radon = Po + He created, does it mean after two half life, radon will be 1/4 x but Po is not 3/4 x (Po will be y and He will be z where y + z = 3/4 x)?
 
  • #10
songoku said:
I am not sure I get what you mean. If the particles lost by radon = Po + He created, does it mean after two half life, radon will be 1/4 x but Po is not 3/4 x (Po will be y and He will be z where y + z = 3/4 x)?
No. Radon decays into Polonium plus Helium.

You can see by the atomic masses than one Polonium plus one Helium almost adds up to one Radon.
 
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  • #11
PeroK said:
Radon decays into Polonium plus Helium plus an alpha particle.
No, the Helium nucleus is the alpha particle - same thing.
 
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  • #12
songoku said:
I am not sure I get what you mean. If the particles lost by radon = Po + He created, does it mean after two half life, radon will be 1/4 x but Po is not 3/4 x (Po will be y and He will be z where y + z = 3/4 x)?
No, I mean that each decaying Radon becomes one Po plus one He, the He being the alpha particle.
 
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  • #13
PeroK said:
No. Radon decays into Polonium plus Helium plus an alpha particle.

You can see by the atomic masses than one Polonium plus one Helium almost adds up to one Radon.

haruspex said:
No, I mean that each decaying Radon becomes one Po plus one He, the He being the alpha particle.

After 1 half life:
Radon = 3/2 × 107 nuclei
Mass of remaining radon = 3/2 × 107 x 222 x 10-3 / (6.02 x 1023) = 5.53 x 10-18 kg

Po = 3/2 × 107 nuclei
Mass of Po = 3/2 × 107 x 218 x 10-3 / (6.02 x 1023) = 5.43 x 10-18 kg

Mass of He = 4 x 1.66 x 10-27 = 6.68 x 10-27 kg

Energy released = (5.43 x 10-18 + 6.68 x 10-27 - 5.53 x 10-18) x (3 x 108)2 = 9 x 10-3 J = 5.625 x 1010 MeV

and it is wrong...:confused:
 
  • #14
songoku said:
After 1 half life:
Radon = 3/2 × 107 nuclei
Mass of remaining radon = 3/2 × 107 x 222 x 10-3 / (6.02 x 1023) = 5.53 x 10-18 kg

Po = 3/2 × 107 nuclei
Mass of Po = 3/2 × 107 x 218 x 10-3 / (6.02 x 1023) = 5.43 x 10-18 kg

Mass of He = 4 x 1.66 x 10-27 = 6.68 x 10-27 kg

Energy released = (5.43 x 10-18 + 6.68 x 10-27 - 5.53 x 10-18) x (3 x 108)2 = 9 x 10-3 J = 5.625 x 1010 MeV

and it is wrong...:confused:

The question specifies 3 half lives. Also, why use kilograms and Joules? That's just making the calculations much more complicated.

Also, surely it's much simpler to calculate how much energy (in ##u## and ##MeV##) is released by one decaying Radon. Then work out how many Radon nuclei decay and multiply the two together.
 
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  • #15
PeroK said:
The question specifies 3 half lives.
I mean my calculation for just 1 half life already wrong, not to mention it still needed to be added with 2 and 3 half life.

Also, why use kilograms and Joules? That's just making the calculations much more complicated.
Just habit to use SI unit

Also, surely it's much simpler to calculate how much energy (in ##u## and ##MeV##) is released by one decaying Radon. Then work out how many Radon nuclei decay and multiply the two together.
Energy released by one decaying Radon = (218.00896 u + 4.00260 u - 222.0157 u) x 931.5 MeV = 3.86 MeV

After three half lives 7/8 of the Radon will have decayed, so total energy released = 3.86 MeV x 7/8 x 3 x 107 = 1.013 x 108 MeV

Is this correct?

If the question asks for only 1 half life, instead of 7/8 I should use 1/2?

Thanks
 
  • #16
songoku said:
Energy released by one decaying Radon = (218.00896 u + 4.00260 u - 222.0157 u) x 931.5 MeV = 3.86 MeV

After three half lives 7/8 of the Radon will have decayed, so total energy released = 3.86 MeV x 7/8 x 3 x 107 = 1.013 x 108 MeV

Is this correct?

Yes.

songoku said:
If the question asks for only 1 half life, instead of 7/8 I should use 1/2?

Yes.
 
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  • #17
Thank you very much for the help Perok and haruspex
 

Related to Energy released through alpha decay

1. What is alpha decay?

Alpha decay is a type of radioactive decay in which an alpha particle (a helium nucleus) is emitted from the nucleus of an atom. This process reduces the atomic number by 2 and the atomic mass by 4.

2. How is energy released through alpha decay?

Energy is released through alpha decay as a result of the conversion of mass into energy. This conversion is based on Einstein's famous equation E=mc^2, where E represents energy, m represents mass, and c represents the speed of light. The energy released is in the form of kinetic energy of the alpha particle and gamma rays.

3. What factors affect the amount of energy released through alpha decay?

The amount of energy released through alpha decay is affected by the mass and stability of the nucleus. Heavier and less stable nuclei tend to release more energy through alpha decay compared to lighter and more stable nuclei.

4. How is the energy released through alpha decay used in practical applications?

The energy released through alpha decay is used in a variety of practical applications, including smoke detectors, nuclear power plants, and cancer treatment. In smoke detectors, the alpha particles are used to ionize air particles, creating an electrical current that triggers an alarm. In nuclear power plants, the energy released is used to generate electricity. In cancer treatment, alpha particles are used to specifically target and destroy cancer cells.

5. What are the potential dangers associated with alpha decay?

Alpha decay can be dangerous if exposure to alpha particles is not properly controlled. These particles can cause damage to living tissue and increase the risk of developing cancer. However, alpha particles can be easily shielded by even a sheet of paper, so the risk is minimal with proper safety precautions in place.

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