Energy released through alpha decay

[songoku]

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 T_1/2 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
T_1/2 = (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

 

[PeroK]

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 T_1/2 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
T_1/2 = (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?

 

[songoku]

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

 

[PeroK]

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!

 

[songoku]

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

 

[PeroK]

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!

 

[songoku]

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

 

[haruspex]

After 2 half life, radon is 1/4 x and Po is 3/4 x. Is this correct?

Yes.

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).

 

[songoku]

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)?

 

[PeroK]

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.

 

[haruspex]

Radon decays into Polonium plus Helium plus an alpha particle.

No, the Helium nucleus is the alpha particle - same thing.

 

[haruspex]

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.

 

[songoku]

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.

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...

 

[PeroK]

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...

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.

 

[songoku]

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

 

[PeroK]

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.

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

Yes.

 

[songoku]

Thank you very much for the help Perok and haruspex