An element $_{x} A^{y}$ emits $5 α$ and $4 β$ particles to give $_{82} B^{207}$ . The number of protons and neutrons in A are respectively.

88, 227

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88, 139

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82, 227

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84, 139

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Solution

The correct option is D $88, 139$
$_{x} A^{y} \to_{82} B^{207} + 5_{2} H e^{4} + 4_{- 1} β^{0}$
On comparing, $x = 82 + 5 \times 2 + 4 (- 1) = 88$
$y = 207 + 5 \times 4 + 4 \times 0$
$= 227$
In $_{88} A^{227}$ ,
the number of protons $= 88$
The number of neutrons $= 227 - 88$
$= 139$

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An element xAy emits 5α and 4β particles to give 82B207. The number of protons and neutrons in A are respectively.

The correct option is D 88,139xAy→82B207+52He4+4−1β0On comparing, x=82+5×2+4(−1)=88y=207+5×4+4×0=227In 88A227,the number of protons =88The number of neutrons =227−88=139

An element $_{x} A^{y}$ emits $5 α$ and $4 β$ particles to give $_{82} B^{207}$ . The number of protons and neutrons in A are respectively.

The correct option is D $88, 139$
$_{x} A^{y} \to_{82} B^{207} + 5_{2} H e^{4} + 4_{- 1} β^{0}$
On comparing, $x = 82 + 5 \times 2 + 4 (- 1) = 88$
$y = 207 + 5 \times 4 + 4 \times 0$
$= 227$
In $_{88} A^{227}$ ,
the number of protons $= 88$
The number of neutrons $= 227 - 88$
$= 139$