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Radioactive polonium 214Po, decays by alpha emission to
1. 214Po(Z=83)
2. 210Pb(Z=82)
3. 214At(Z=85)
4. 210Bi(Z=83)

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Correct Answer - Option 2 : 210Pb(Z=82)

CONCEPT:

Radioactivity:

  • Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive.
  • radioactive nucleus consists of an unstable assembly of protons and neutrons which becomes more stable by emitting an alpha, a beta particle, or a gamma photon.
  • Atoms are radioactive if their nuclei are unstable and spontaneously (and random) emit various particles α, β, and/or γ radiations.
  • Three crucial forms of Radioactivity:
    • Gamma Decay-(Photons having high energy are throw down).
    • Beta Decay-(Discharge consists of Electrons).
    • Alpha Decay-(Discharge consists of a Helium nucleus).
  • An α particle is essentially a He nucleus consisting of two protons and two neutrons.
  • Two protons correspond to 2 units of charge on an α particle.
  • Two protons and two neutrons correspond to 4 units of mass of an α particle.  
  • Hence, the symbol for an α particle is 4He2.
  • From the definition of α particle given above, emission of an α particle will lead to a decrease in 2 units of charge and a decrease in 4 units of mass of the atom.

CALCULATION:

Given Atomic number Z1 = 84 and Mass number A1 = 214

  • An α particle is essentially a He nucleus consisting of two protons and two neutrons.
  • From the definition of α particle given above, emission of an α particle will lead to a decrease in 2 units of atomic number and a decrease in 4 units of mass of the atom.

So when the polonium 214Po, decays by alpha emission, then the resultant element will have,

⇒ Z2 = Z1 - 2

⇒ Z2 = 84 - 2

⇒ Z2 = 82

And,

⇒ A2 = A1 - 4

⇒ A2 = 214 - 4

⇒ A2 = 210

The Pb (Lead) has the atomic number 82. Therefore the new element will be,

⇒ 210Pb(Z=82)

  • Hence, option 2 is correct.

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