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Standard X

Physics

Alpha Decay

Write down th...

Question

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

α - γ - β

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α - β - γ

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γ - β - α

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β - γ - μ

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Solution

The correct option is A $α - γ - β$
$_{Z + 1} X^{A} \to_{Z - 1} Y^{A - 4} +_{2} α^{4} \to_{Z - 1} Y^{A - 4} +_{0} γ^{0} \to_{Z} K^{A - 4} +_{- 1}^{0} β$
So, sequence is $α \to γ \to β$
So, (A) is correct option

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Q. Write down the sequence of radiations emitted in the following radioactive processe.

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

Q. Let

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β = 2 α

and

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. Then the value of

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

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_{z} X^{A}

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

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Q. If

\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

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 $α$ , $β$ , & $γ$ .

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Write down the sequence of radiations emitted in the following radioactive processes.z+1XA→z−1YA−4→z−1YA−4→zKA−4

The correct option is A α−γ−βZ+1XA→Z−1YA−4+2α4→Z−1YA−4+0γ0→ZKA−4+0−1βSo, sequence is α→γ→βSo, (A) is correct option

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

The correct option is A $α - γ - β$
$_{Z + 1} X^{A} \to_{Z - 1} Y^{A - 4} +_{2} α^{4} \to_{Z - 1} Y^{A - 4} +_{0} γ^{0} \to_{Z} K^{A - 4} +_{- 1}^{0} β$
So, sequence is $α \to γ \to β$
So, (A) is correct option