Alpha Decay of Polonium-210 Homework

In summary: Pb emits alpha radiation, which will cause other materials to become radioactive if they come into contact with it. However, Po cannot cause cancer directly, only indirectly.
  • #1
Lissajoux
82
0

Homework Statement



Polonium isotope [itex]{}_{84}^{210}Po[/itex] is unstable and emits a 5.30MeV [itex]\alpha[/itex] particle.

i. Indentify the daughter nuclide (using a periodic table)

ii. Calculate the mass of the daughter nuclide.

Homework Equations



Mass of neutral [itex]{}^{210}Po[/itex] nuclide is 209.9829u

Mass of [itex]{}^{4}He[/itex] is 4.0026u

Note that taking the mass of the alpha particle as the mass of a neutral helium atom.

The Attempt at a Solution



i. [tex]{}_{84}^{210}Po \rightarrow {}_{82}^{206}? + {}_{2}^{4}\alpha + Q[/itex]

By looking at a periodic table, identified the daughter nuclide as: [itex]{}_{82}^{206}Pb[/itex]

.. correct?

ii. Not sure how to do this. The mass of both sides of the equation must be the same I assume. Already know the mass of the alpha particle, it's the mass of the helium atom given. The mass of the neutral Po I assumed would refer to the nuclide after the alpha particle has been emitted, but this is Pb not Po.

Not sure how to get the values required to perform the calculation, which I guess would be:

[tex]M({}_{84}^{210}Po) - M({}_{2}^{4}\alpha) = M({}_{84}^{210}Pb)[/itex]

So 2 things with this part, how to I get the values of the masses? and is the equation to calculate the mass of the daughter nuclide correct?

Any help really will be appreciated.
 
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  • #2
Your last equation is partially correct. You also need to subtract off the kinetic energy of the alpha particle.
 
  • #3
Is it correct to do this:

Mass of neutral [itex]{}^{210}Po[/itex] nuclide is 209.9829u

Mass of [itex]{}^{4}He[/itex] is 4.0026u

[tex]M({}_{84}^{210}Pb)=M({}_{84}^{210}Po) - M({}_{2}^{4}\alpha)=205.98u=0.19MeV[/tex]

Not sure if it's that simple? If not, where am I going wrong?
 
  • #4
The masses were given to you. So those should be correct. Also it asks for the mass of Pb, so you should keep it in atomic units and not MeV. Also, as I said before, that is not correct. You still need to subtract off the kinetic energy of the alpha particle.
 
  • #5
How do I calculate the kinetic energy of the alpha particle in order to then subtract this?
 
  • #6
They give it to you already in the problem. (5.30MeV)
 
  • #7
Right, Ok.

So if I change the equation to this:

[tex]
M({}_{84}^{210}Pb)=M({}_{84}^{210}Po) - M({}_{2}^{4}\alpha)-Q
[/tex]

Then I either take off that Q=5.30MeV from my answer then convert it from MeV to u, or convert 5.30MeV to u and then take it away from my answer given in u. If that makes sense.
 
  • #8
Do the latter part. Try to keep it in u's, so your final answer will be a mass and not energy.
 
  • #9
When Po is injested into the body. It emits alpha radiation. Does this radiation causes other materials in the body to become radioactive?
 

What is alpha decay?

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, consisting of two protons and two neutrons, from its nucleus. This process reduces the atomic number of the nucleus by two and the mass number by four.

What is Polonium-210?

Polonium-210 is a radioactive isotope of the chemical element polonium. It has a half-life of 138 days and undergoes alpha decay to form lead-206. It is commonly used in static eliminators, nuclear batteries, and as a heat source in space probes.

What is the purpose of studying alpha decay of Polonium-210?

Studying alpha decay of Polonium-210 is important in understanding the behavior and characteristics of this radioactive isotope. It also has practical applications, such as in the production of static electricity and generating heat for space missions.

How is alpha decay of Polonium-210 calculated?

The rate of alpha decay of Polonium-210 can be calculated using the radioactive decay equation, where the rate of decay is equal to the initial amount of Polonium-210 multiplied by the decay constant and the time elapsed. This calculation can help determine the amount of Polonium-210 present at a given time.

What are the potential hazards of working with Polonium-210?

Polonium-210 is highly radioactive and emits alpha particles, which can be harmful to living organisms. Exposure to large amounts of Polonium-210 can lead to radiation sickness and an increased risk of cancer. It is important to handle this isotope with proper safety precautions and to limit exposure to it.

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