Atomic mass number of an element is 232 and its atomic number is 90. The end product of this radioactive element is an isotope of lead (atomic mass 208 and atomic number 82). The number of α-and β-particles emitted are

Question

Atomic mass number of an element is 232 and its atomic number is 90. The end product of this radioactive element is an isotope of lead (atomic mass 208 and atomic number 82). The number of α-and β-particles emitted are (A) α=3, β=3 (B) α=6, β=4 (C) α=6, β=0 (D) α=4, β=6

Tardigrade Admin · Accepted Answer

Number of α-particles emitted =(232-208/4)=6 Decrease in charge number due to α-emission =12. But actual decrease in charge number =90-82=8. Clearly, four β-particles are emitted.

Q. Atomic mass number of an element is $232$ and its atomic number is $90$ . The end product of this radioactive element is an isotope of lead (atomic mass $208$ and atomic number $82)$ . The number of $α$ -and $β$ -particles emitted are

$α = 3, β = 3$

$α = 6, β = 4$

$α = 6, β = 0$

$α = 4, β = 6$

Number of $α$ -particles emitted $= \frac{232 - 208}{4} = 6$
Decrease in charge number due to $α$ -emission $= 12$ .
But actual decrease in charge number $= 90 - 82 = 8$ .
Clearly, four $β$ -particles are emitted.

Q. Atomic mass number of an element is 232 and its atomic number is 90. The end product of this radioactive element is an isotope of lead (atomic mass 208 and atomic number 82). The number of α-and β-particles emitted are

α=3,β=3

α=6,β=4

α=6,β=0

α=4,β=6