How would you write a balanced nuclear equations for the emission of an alpha particle by polonium-209 and for the decay of calcium-45 to scandium-45?

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How would you write a balanced nuclear equations for the emission of an alpha particle by polonium-209 and for the decay of calcium-45 to scandium-45?

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Answer 1 · 2015-12-05T13:56:25

$_{84}^{209} Po \to_{82}^{205} Pb +_{2}^{4} α$ $""_84^209"Po" -> ""_82^205"Pb" + ""_2^4alpha$

$_{20}^{45} Ca \to_{21}^{45} Sc +_{-1}^{0} e +_{0}^{0} ν$ $""_20^45"Ca" -> ""_21^45"Sc" + ""_text(-1)^0"e" + ""_0^0nu$

Explanation:

The alpha decay of polonum-209, which consists of the emission of an alpha particle, will leave behind the atom of a different element.

Notice that the identity of the element changes after the decay. This means that you're dealing with a nuclear transmutation, a process that converts an atom of a given element into an atom of a different element.

Now, an alpha particle consists of

two protons

two neutrons

In essence, an alpha particle is a helium-4 nucleus, which means that you can see it as having anatomic number equal to $2$ $2$ and a mass number equal to $4$ $4$ .

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When a nucleus emits an alpha particle, its atomic number will decrease by $2$ $2$ , since it's losing two protons, and its mass number will decrease by $4$ $4$ , since it's losing two protons and two neutrons.

You can thus say that

$_{84}^{209} Po \to_{Z}^{A} X +_{2}^{4} α$ $""_84^209"Po" -> ""_Z^A"X" + ""_2^4alpha$

Balance the nuclear equation by balancing the atomic number and the mass number for both sides of the equation

$209 = A + 4$ $209 = A + 4 " "$ and $84 = Z + 2$ $" "84 = Z + 2$

The atomic number and the mass number of the resulting element will be

$A = 209 - 4 = 205$ $A = 209 - 4 = 205" "$ and $Z = 84 - 2 = 82$ $" "Z = 84 - 2 = 82$

A quick look in the periodic table will show that you're dealing with lead-205

$_{84}^{209} Po \to_{82}^{205} Pb +_{2}^{4} α$ $""_84^209"Po" -> ""_82^205"Pb" + ""_2^4alpha$

Now for the decay of calcium-45 to scandium-45. Two important things to notice here

you're once again dealing with a nuclear transmutation

the mass number is conserved following the decay

Turn to the periodic table again and make a note of the atomic numbers of the two elements

$Z_{C a} = 20$ $Z_(Ca) = 20 " "$ and $Z_{S c} = 21$ $" "Z_(Sc) = 21$

This means that you're looking for a nuclear decay that will increase the atomic number by $1$ $1$ and keep the mass number constant.

As you know, in beta decay, a neutron is being converted into a proton and a beta particle and an antineutrino are being emitted from the nucleus.

A beta particle is simply an electron $β =_{-1}^{0} e$ $beta = ""_text(-1)^0"e"$

This means that you will have

$_{20}^{45} Ca \to_{21}^{45} Sc +_{-1}^{0} e +_{0}^{0} ν$ $""_20^45"Ca" -> ""_21^45"Sc" + ""_text(-1)^0"e" + ""_0^0nu$

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How would you write a balanced nuclear equations for the emission of an alpha particle by polonium-209 and for the decay of calcium-45 to scandium-45?

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