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Alpha Decay

In a given re...

Question

In a given reaction
$_{Z} A^{A} ⟶_{Z + 1} Y^{A} ⟶_{Z - 1} K^{A - 4} ⟶_{Z - 1} K^{A - 4}$
Radioactive radiations are emitted in the sequence of

A

α

,

β

,

γ

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B

γ

,

α

,

β

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C

β

,

α

,

γ

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D

γ

,

β

,

α

,

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Solution

The correct option is C $β$ , $α$ , $γ$
Alpha particle is an helium atom and contains two neutrons and two protons.
The beta particle is an energetic electron given off by the nucleus of unstable isotopes to restore an energy balance gamma rays are energetic photon.

Therefore we can observe the order of particle emission as $β, α, γ$ .

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Similar questions

z

be a imaginary number satisfying

| z - 1 | = 1

α = 2 z

β = 2 α

γ = 2 β

. Then the value of

{| z |}^{2} + {| α |}^{2} + {| β |}^{2} + {| γ |}^{2} + {| z - 2 |}^{2} + {| α - 4 |}^{2} + {| β - 8 |}^{2} + {| γ - 16 |}^{2}

\frac{x}{tan (θ + α)} = \frac{y}{tan (θ + β)} = \frac{z}{tan (θ + γ)}

then prove that

\frac{x + y}{x - y} {sin}^{2} (α - β) + \frac{y + z}{y - z} {sin}^{2} (β - γ) + \frac{z + x}{z - x} {sin}^{2} (γ - α) = 0

Q. In the given reaction

_{z} X^{A} \to_{z + 1} Y^{A} \to_{z - 1} K^{A - 4} \to_{z - 1} K^{A - 4}

Radioactive radiations are emitted in the sequence of

In each of $α$ $β$ and $γ$ $is a positive acute angle such that$
$sin (α + β - γ) = \frac{1}{2}, cos (β + γ - α) = \frac{1}{2} and tan (γ + α - β) = 1$
Find the values of $α$ , $β$ , & $γ$ .

Q. Show that the lines

\frac{x - a + d}{α - δ} = \frac{y - a}{α} = \frac{z - a - d}{α + δ}

\frac{x - b + c}{β - γ} = \frac{y - b}{β} = \frac{z - b - c}{β + γ}

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