Question

What is the final nuclide that results from a succession of nine alpha emissions, followed by six beta emissions, beginning with Plutonium-244, 244Pu? 226 Gd 202Pb 208 po 208Pb 208Yb

Answer 1

In Nuclear chemistry $\alpha$ decay or emission from one radioactive element produces another element which is two places to the left side in the periodic table.

$\alpha$ particle = isotope

He

X ------- > Y + 9 $\alpha$

New element after 9 $\alpha$ decays = Y

1$\alpha$ decay produce a new element which is 2 places left to the original element ( X here)

So , Nine $\alpha$ decays produce a new element which is 2 places( atomic number is 2 less than the first one) left to the original element ( X here)

SO Atomic number of Pu = 94

Atomic number  of new element = 94 - 9 X2

= 76

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And $\beta$ decay or emission from one radioactive element produces another element which is one place to the right side in the periodic table.

Six $\beta$ decays produces a new element six places right to the first element.

So, 76 + 6 X1 = 82

So , Atomic number 82 is lead , Pb

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Now there are two lead isotpes in the given options.

Mass number to be identified.

Now let us determine the final nuclide after 9 $\alpha$ particles and 6 $\beta$ pparticles decays from the given 244-Pu nuclide.

Plutonium isotope------

94 244 Pu
--------------> Y + 9
He

Y =?

Atomic number = 94 - 9 * 2

= 76

mass number = 244 - 9 * 4 ( mass n umber of helium nuclide is 4)

= 208

So Y = $_{208}^{76}\textrm{Os}$ '

Now 6 beta decyas from $_{208}^{76}\textrm{Os}$

Final nuclide = $_{208}^{76}\textrm{Os}$ + 6 beta particles ( beta particle is $_{-1}^{0}\textrm{e}$ )

The final nuclide is Six places right to osmium .

new atomic nuumber is 76 + 7 = 82 i.e lead.

mass number of Pb isotope = 208

There will be no mass number change in beta decay.

The final nuclide = $_{82}^{208}\textrm{Pb}$