A decay is given by 1532 P -1632 S - A - B- then A - B areA- -1- v antineutrinoB- -1- gamma particleC- - v antineutrino D- -1- v neutrino

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Rutherford's Model

A decay is gi...

Question

A decay is given by $_{15}^{32} P ⟶_{16}^{32} S + A + B$ , then A & B are

β^{- 1}

, gamma particle

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β^{- 1}, \to v

(antineutrino)

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α, \to v

(antineutrino)

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β^{- 1}, v

(neutrino)

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Solution

The correct option is B $β^{- 1}, \to v$ (antineutrino)
Beta decay is a process in which unstable atoms undergoes radioactivity to become more stable.
There are two types of beta decay, beta-minus and beta-plus. During beta-minus decay, a neutron in an atom's nucleus turns into a proton, an electron and an antineutrino. The electron and antineutrino fly away from the nucleus, which now has one more proton than it started with. Since an atom gains a proton during beta-minus decay, it changes from one element to another.

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A decay is given by 3215P⟶3216S+A+B, then A & B are

A decay is given by $_{15}^{32} P ⟶_{16}^{32} S + A + B$ , then A & B are