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# For the following nuclear disintegration process${}_{92}^{238}\text{U }\to \text{ }{}_{82}^{206}\text{Pb + }x\text{ }\!\![\!\!\text{ }{}_{2}^{4}\text{He }\!\!]\!\!\text{ + 6 }\!\![\!\!\text{ }{}_{-1}^{0}e]$. The value of x is – A) 10B) 4C) 6D) 8

Last updated date: 16th Sep 2024
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Hint: In a disintegration process, the total mass in a reaction is always conserved. We can tally the mass numbers of the reactants and the products to find any missing value. An electron's only charge balance doesn’t count to the mass.

Complete step-by-step solution
An unstable and heavy atom can undergo disintegration due to various reasons. A Uranium – 238 atom is a typical example of which undergoes almost any form of disintegration such as alpha decay. The U-238 attains stability by becoming a nucleus of another light-weight atom. During this process, there will be the emission of energy along with other by-products like an alpha particle, an electron, or a positron.
Now, let us consider the disintegration process given to us –
${}_{92}^{238}\text{U }\to \text{ }{}_{82}^{206}\text{Pb + }x\text{ }\!\![\!\!\text{ }{}_{2}^{4}\text{He }\!\!]\!\!\text{ + 6 }\!\![\!\!\text{ }{}_{-1}^{0}e]$
In this process a U-238 atom is disintegrated to one Pb-206 atom, 6 electrons and ‘x’ number of helium atoms.
We need to find the number of helium atoms involved in the disintegration of a single nucleus of U-238.
For this, we can simply equate the mass numbers of either side of the balanced equation.
Assuming that the given equation is balanced, we can equate the atomic masses on both sides as:
\begin{align} & 238=206+4x \\ & \Rightarrow \text{ }x=\dfrac{238-206}{4} \\ & \Rightarrow \text{ }x=8 \\ \end{align}
So, the number of Helium atoms produced from the disintegration of a single U-238 nucleus is 8.
The correct answer is option D.