Write the symbols and a balanced nuclear equation for the following:

a. Polonium -210 decays to give lead -206

b. Bismuth -211 decays by emitting an alpha particle

c. The product from b. emits a beta particle

d. When an alpha particle bombards aluminum -27, one product is silicon -30

I'll do the first one for you. The three things you keep in mind when doing these problems is

1. The subscripts must add up on the left and right.
2. The superscripts must add up on the left and right.
3. The identify of the element (X for example) is determined by the atomic number which you can find on the periodic table. (You can't find beta + or beta - or protons on the periodic table since these are not elements; you must know what those are)

The superscripts are mass numbers, the subscripts are atomic numbers.

21084Po ==>20682Pb + X
All of the above is in the problem. All you need to do is to identify X.
210 mass number on the left and 206 on the right; therefore, X must have mass number of 4. Atomic number of 84 on the left and 82 on the right; therefore, the atomic number must be 2. Look on the periodic table and atomic number 2 is He; therefore, we write X as 42He
The others are done the same way; keep in mind charge and mass for beta +, beta -, protons, neutrons, etc.

a. Polonium-210 decays to give lead-206.

To write a balanced nuclear equation for this decay, we need to understand that the sum of atomic numbers and mass numbers must be equal on both sides of the equation.

Polonium-210 has an atomic number of 84 and a mass number of 210, while lead-206 has an atomic number of 82 and a mass number of 206.

The decay process for polonium-210 can be represented as:

^210Po -> ^206Pb + X

The symbol "X" represents the emission of a particle (such as an alpha particle, a beta particle, or a gamma ray) during the decay. In this case, no particle emission is specified, so we can assume it is omitted.

However, the equation is not balanced yet. To balance it, we need to ensure that the atomic number and the mass number are equal on both sides.

^210Po -> ^206Pb + ^4He

This balanced equation represents the decay of polonium-210 into lead-206, with the emission of an alpha particle (^4He).

b. Bismuth-211 decays by emitting an alpha particle.

Similar to the previous case, we want to write a balanced nuclear equation:

^211Bi -> Y + ^4He

Here, "Y" represents an unknown product. The decay of bismuth-211 by emitting an alpha particle is represented by the balanced equation above.

c. The product from b. emits a beta particle.

To represent this process, we can use the balanced equation from part b, as the product from part b undergoes another decay step:

Y -> Z + e^- + v

In this equation, "Z" represents the unknown final product, "e^-" denotes an electron (beta particle), and "v" is a symbol for an electron antineutrino.

d. When an alpha particle bombards aluminum-27, one product is silicon-30.

To write a balanced nuclear equation for this reaction, we need to consider that an alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons.

^4He + ^27Al -> ^30Si + X

In this equation, "X" represents an unknown particle emitted during the reaction. The equation represents the bombardment of aluminum-27 with an alpha particle, resulting in the formation of silicon-30 as one of the products.

a. Polonium-210 decays to give lead-206

Symbol:
Polonium-210 -> Lead-206

Balanced nuclear equation:
^210Po -> ^206Pb + ^4He

b. Bismuth-211 decays by emitting an alpha particle

Symbol:
Bismuth-211 -> Alpha particle + Unknown element

Balanced nuclear equation:
^211Bi -> ^4He + Unknown element

c. The product from b. emits a beta particle

Symbol:
Unknown element -> Beta particle + Unknown element 2

Balanced nuclear equation:
Unknown element -> ^0β + Unknown element 2

d. When an alpha particle bombards aluminum-27, one product is silicon-30

Symbol:
Alpha particle + Aluminum-27 -> Silicon-30 + Unknown particle

Balanced nuclear equation:
^4He + ^27Al -> ^30Si + Unknown particle