Question

How many $\alpha$-particle and $\beta$-particles are emitted when uranium nucleus ${ }_{92}^{238} U$ decays to lead nucleus ${ }_{82}^{206} P b$ ?

• A. $ \alpha =6,\beta =8 $
• B. $ \alpha =10,\beta =8 $
• C. $ \alpha =8,\beta =10 $
• D. $ \alpha =8,\beta =6 $

Answer 1

The Correct Option is D

Let $x$ number of $\alpha$-particles and $y$ number of $\beta$-particles be emitted, when uranium nucleus decays to lead nucleus. ${ }_{92}^{238} U \rightarrow{ }_{82}^{206} P b+x(\alpha)+y(\beta)$ Since, $\alpha$-particle is doubly ionized helium atom and $\beta$ - particles are fast moving electrons, also mass number and atomic number remains conserved in the reaction. $\therefore{ }_{92}^{238} U \rightarrow{ }_{82}^{206} P b+x\left({ }_{2} H e^{4}\right)+y\left({ }_{1} \beta^{0}\right)$ Equating mass number, we have $238=206+4 x$ $\Rightarrow x=8$ Equating atomic number, we have $92=82+2 x-y=82+(2 \times 8)-y$ $\Rightarrow y=6$ Hence, $\alpha=8, \beta=6$.