Write down the sequence of radiations emitted in the following radioactive processes.
$_{z + 2} X^{A} \to_{z} Y^{A - 4} \to_{z + 1} Y^{A - 4} \to_{z - 1} K^{A - 8}$

γ - β - α

No worries! We‘ve got your back. Try BYJU‘S free classes today!

γ - α - β

No worries! We‘ve got your back. Try BYJU‘S free classes today!

β - γ - α

No worries! We‘ve got your back. Try BYJU‘S free classes today!

α - β - α

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $α - β - α$
$_{Z + 2} X^{A} \to_{Z} Y^{A - 4} +_{2} α^{4} \to_{Z + 1} Y^{A - 4} +_{+ 1}^{0} β \to_{Z - 1} K^{A - 8} +_{2} α^{4}$

So, $α \to β^{+} \to α$
So, (D) is correct option.

Suggest Corrections

Similar questions

z

| z - 1 | = 1

α = 2 z

β = 2 α

γ = 2 β

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

_{z} X^{A}

_{z - 1} Y^{A - 4}

α

β

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

\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

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

_{z} X^{A} \to_{z} P^{A} \to_{z - 2} Y^{A - 4} \to_{z - 1} Y^{A - 4}

Join BYJU'S Learning Program