The radioactive decay of uranium into thorium is expressed by the equation
$_{92} U^{238} \to_{90} T h^{234} + X$ Where X is

an electron

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a proton

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a deuteron

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an alpha particle

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Solution

The correct option is D an alpha particle
$_{92} U^{238} \to_{90} T h^{234} + X$
Total atomic number and mass number should be equal to both sides of equation.
Atomic number of $X = 92 - 90 = 2$
Mass number of $X = 238 - 234 = 4$
So , element is $_{2} X^{4}$ , hence it is $α$ particle.( Because atomic number of $α$ is $2$ and mass number is $4$ )

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Similar questions

_{92} U^{238}

238.0508 u

_{90} {Th}^{234}

234.0436 u

_{2} {He}^{4}

4.0026 u

α - decay

_{92} U^{238}

_{90} {Th}^{234}

[1 u \times c^{2} = 931.5 MeV]

The radioactive decay of uranium into thorium is represented by the equation
$_{92}^{238} U \to_{90}^{234} T h + x$
What is x?

_{92} U^{238} \to_{90} T h^{234} \to_{91} P a^{234} \to

_{92} U^{234} \to_{90} T h^{230} \to_{88} R a^{226}

_{92} U^{238}

_{82} {P b}^{206}

α

β

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