Question

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

• A. $88, 227$
• B. $88, 139$
• C. $82, 227$
• D. $84, 139$

Answer 1

Answer: (B)

$88, 139$

Answer 2

${ }_{x} A^{y} \to { }_{82} B^{207}+5_{2} H e^{4}+4_{-1} \beta^{0}$

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 $=227-88$
$=139$