What is the respective number of α and β particles emitted in the following radioactive decay ? 90 X 200 longrightarrow 80 Y168

Question

What is the respective number of α and β particles emitted in the following radioactive decay ? 90 X 200 longrightarrow 80 Y168 (A) 6 and 8 (B) 8 and 8 (C) 6 and 6 (D) 8 and 6

Tardigrade Admin · Accepted Answer

nα=(A-A'/4) =(200-168/4) =8 nβ =2 nα-Z+Z' =2 × 8-90+80=6

Q. What is the respective number of $α$ and $β$ particles emitted in the following radioactive decay ?
$_{90} X^{200} ⟶_{80} Y^{168}$

6 and 8

8 and 8

6 and 6

8 and 6

$n_{α} = \frac{A - A ^{'}}{4}$
$= \frac{200 - 168}{4}$
$= 8$
$n_{β} = 2 n_{α} - Z + Z^{'}$
$= 2 \times 8 - 90 + 80 = 6$

Q. What is the respective number of α and β particles emitted in the following radioactive decay ? 90​X200⟶80​Y168