When 90 T h 228 transforms to 83 Bi 212- then the number of the emitted -and -particles is- respectivelyA- 8 - 7 - 4 - 7 -C- 4 - 4 -D- 4 - 1 -

The correct option is D 4α , 1β_Z=90Th^A=228→ _Z=83Bi^A'=212 Number of α particles emitted n_α =A A'/4=228 212/4=4 Number of β particles emitted n_β =2n_α Z ...

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Standard X

Physics

Beta Decay

When 90 T h 2...

Question

When $_{90} T h^{228}$ transforms to $_{83} B i^{212}$ , then the number of the emitted $α$ -and $β$ -particles is, respectively

8 α, 7 β

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4 α, 7 β

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4 α, 4 β

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4 α, 1 β

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Solution

The correct option is D $4 α, 1 β$
$_{Z = 90} T h^{A = 228} \to_{Z = 83} B i^{A^{'} = 212}$
Number of $α$ -particles emitted $n_{α} = \frac{A - A^{'}}{4} = \frac{228 - 212}{4} = 4$
Number of $β$ -particles emitted $n_{β} = 2 n_{α} - Z + Z^{'} = 2 \times 4 - 90 + 83 = 1$

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Similar questions

Q. When

_{90} T h^{228}

transforms to

_{83} B i^{212}

, then the number of the emitted

α

-and

β

-particles is, respectively

Q. When

_{90} {Th}^{228}

gets converted into

_{83} {Bi}^{212}

, the number of

α

and

β

particles emitted will respectively be-

Q. The number of

α

and

β^{-}

particles emitted in nuclear reaction,

_{90} T h^{228} ⟶_{83} B i^{212}

are respectively :

t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α

$t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α$ =

[IIT 1988; MP PET 1991]

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When 90Th228 transforms to 83Bi212, then the number of the emitted α -and β -particles is, respectively

The correct option is D 4α,1βZ=90ThA=228→Z=83BiA′=212 Number of α -particles emitted nα=A−A′4=228−2124=4 Number of β -particles emitted nβ=2nα−Z+Z′=2×4−90+83=1

When $_{90} T h^{228}$ transforms to $_{83} B i^{212}$ , then the number of the emitted $α$ -and $β$ -particles is, respectively

The correct option is D $4 α, 1 β$
$_{Z = 90} T h^{A = 228} \to_{Z = 83} B i^{A^{'} = 212}$
Number of $α$ -particles emitted $n_{α} = \frac{A - A^{'}}{4} = \frac{228 - 212}{4} = 4$
Number of $β$ -particles emitted $n_{β} = 2 n_{α} - Z + Z^{'} = 2 \times 4 - 90 + 83 = 1$