# In a radioactive reaction $_{92}{X^{232}} \to$$_{82}{Y^{204}}$, the number of $\alpha$- particles emitted isA. 7B. 6C. 5D. 4

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Hint:Alpha decay is a type of decay in which a nucleus transforms itself into a different nucleus by emission of alpha particles. Alpha particles$\left( \alpha \right)$is a helium nucleus. The$\alpha$decay is shown as$_A{X^Z}{ \overset{\alpha }{\rightarrow} _{A - 2}}{Y^{Z - 4}}$, where$\left( A \right)$is the atomic mass and$\left( Z \right)$is the atomic number. To find the emission of alpha particles subtract the atomic masses of X and Y and divide it by 4.

Complete step by step solution:
From your physics lessons, you have learned about radioactivity,$\alpha$- decay and$\beta$- decay. In alpha decay, it is a type of decay in which a nucleus transforms itself into a different nucleus by the emission of alpha particles. Nucleus of helium $\left( {_2H{e^4}} \right)$ is the alpha particle. Helium contains two neutrons and two protons.

Therefore after the emission, the atomic mass of the nucleus which is going through emission reduces to 4 and the atomic number reduces 2.

Hence the change of$_Z{X^A}$to the emitting nucleus$_{Z - 2}{X^{A - 4}}$is expressed as $_Z{X^A}{ \to _{Z - 2}}{Y^{A - 4}}{ + _2}H{e^4}$

In helium atom$\alpha$- particle is a nucleus. When an alpha particles is emitted from a nucleus, the resultant nucleus reduces in mass number by 4 link and in atomic number by 2 unit. Loss in mass number $= 232 - 204 = 2$

Therefore, number of$\alpha$- particles emitted $=\dfrac{28}{4}=7$

Thus the number of alph$\left( \alpha \right)$ particles emitted is 7.

Hence, The correct option is A.

Note: The reaction that is given about is not a direct decay because it has gone under many alphas decays. There is no requirement of external energy in the emission of alpha particles. The decrease in atomic mass is only caused by alpha decay. The decrease in atomic mass is not caused by beta decay. It can only increase or decrease the atomic number.