
Three $\alpha $-particles and one $\beta $-particle decaying takes place in series from an isotope ${}_{88}R{a^{236}}$. Finally the isotope obtained will be
A) ${}_{84}{X^{220}}$
B) ${}_{86}{X^{222}}$
C) ${}_{83}{X^{224}}$
D) ${}_{83}{X^{215}}$
Answer
162.9k+ views
Hint:The problem is from the radioactive decay section of physics. There are three types of radioactive decay processes. We need to apply the chemical equations of alpha and beta decay to find the resultant isotope.
Complete answer:
The emission of three $\alpha $-particles is given by the chemical equation.
${}_{88}{Ra^{236}}\longrightarrow{}_{82}{X^{224}}+3{}_{2}{He^{4}}$
The emission of one $\beta $-particle is given by the chemical equation.
${}_{82}{X^{224}}\longrightarrow{}_{83}{X^{224}}+{}_{-1}{e^{0}}$
Hence, the correct option is Option (C).
Additional Information:
A radioactive atom will naturally emit radiation in the form of energy or particles in order to transition into a more stable state. This process is called radioactive decay. One element spontaneously changes into another during radioactive decay. Alpha decay, beta decay, and gamma decay are three of the decay types that all involve the emission of one or more particles.
Alpha decay: The nuclear decay process known as "alpha decay" results in the parent nucleus emitting an alpha particle. The alpha particle, which has two protons and two neutrons, is the helium atom's nucleus.
${}_{Z}{X^{A}}\longrightarrow{}_{Z-2}{Y^{A-4}}+{}_{2}{He^{4}}$
Beta decay: A beta particle is released from an atomic nucleus during the radioactive decay process known as beta decay. There are two types of beta decay: negative beta decay which emits electrons and positive beta decay which emits positrons.
${}_{Z}{X^{A}}\longrightarrow{}_{Z+1}{Y^{A}}+{}_{-1}{e^{0}}$
${}_{Z}{X^{A}}\longrightarrow{}_{Z-1}{Y^{A}}+{}_{+1}{e^{0}}$
Gamma decay: Gamma decay is a type of radioactivity where some unstable atomic nuclei release extra energy through electromagnetic radiation. The emission of alpha or beta particles generally leaves the resulting nuclei in an excited state. Then return to the stable ground state is accompanied by the emission of gamma particles (Photons).
$({}_{Z}{X^{A}})^*\longrightarrow{}_{Z}{X^{A}}+\gamma$
Note:While solving this question don’t get confused between beta negative and beta positive decay. Also you need to know about what atomic and mass numbers are.
Complete answer:
The emission of three $\alpha $-particles is given by the chemical equation.
${}_{88}{Ra^{236}}\longrightarrow{}_{82}{X^{224}}+3{}_{2}{He^{4}}$
The emission of one $\beta $-particle is given by the chemical equation.
${}_{82}{X^{224}}\longrightarrow{}_{83}{X^{224}}+{}_{-1}{e^{0}}$
Hence, the correct option is Option (C).
Additional Information:
A radioactive atom will naturally emit radiation in the form of energy or particles in order to transition into a more stable state. This process is called radioactive decay. One element spontaneously changes into another during radioactive decay. Alpha decay, beta decay, and gamma decay are three of the decay types that all involve the emission of one or more particles.
Alpha decay: The nuclear decay process known as "alpha decay" results in the parent nucleus emitting an alpha particle. The alpha particle, which has two protons and two neutrons, is the helium atom's nucleus.
${}_{Z}{X^{A}}\longrightarrow{}_{Z-2}{Y^{A-4}}+{}_{2}{He^{4}}$
Beta decay: A beta particle is released from an atomic nucleus during the radioactive decay process known as beta decay. There are two types of beta decay: negative beta decay which emits electrons and positive beta decay which emits positrons.
${}_{Z}{X^{A}}\longrightarrow{}_{Z+1}{Y^{A}}+{}_{-1}{e^{0}}$
${}_{Z}{X^{A}}\longrightarrow{}_{Z-1}{Y^{A}}+{}_{+1}{e^{0}}$
Gamma decay: Gamma decay is a type of radioactivity where some unstable atomic nuclei release extra energy through electromagnetic radiation. The emission of alpha or beta particles generally leaves the resulting nuclei in an excited state. Then return to the stable ground state is accompanied by the emission of gamma particles (Photons).
$({}_{Z}{X^{A}})^*\longrightarrow{}_{Z}{X^{A}}+\gamma$
Note:While solving this question don’t get confused between beta negative and beta positive decay. Also you need to know about what atomic and mass numbers are.
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