First slide

Radio activity

Question

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

Easy

A

$8 α, 7 β$
B

$4 α, 7 β$
C

$4 α, 4 β$
D

$4 α, 1 β$

Solution

$n_{α} = \frac{228 - 212}{4} = 4 and n_{β} = 2 x 4 - 90 + 83 = 1$

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