Question

( a ) Calculate number of $\alpha$ and $\beta -$ particles emitted when ${}_{92}{U}^{238}$ changes into radioactive ${}_{82}P{b}^{206}.$
( b )$T{h}^{234}$ disintegrates and emits $6\beta -$ and $7\alpha -$ particles to form a stable element. Find the atomic number and
mass number of the stable product. Also identify the element:

• A. ( a ) No. of $\alpha -$ particles$=8$, No. of $\beta -$ particles$=6$; ( b ) ${}_{82}P{b}^{206}$
• B. ( a ) No. of $\alpha -$ particles$=6$, No. of $\beta -$ particles$=8$; ( b ) ${}_{82}P{b}^{206}$
• C. ( a ) No. of $\alpha -$ particles$=6$, No. of $\beta -$ particles$=8$; ( b ) ${}_{92}P{b}^{206}$
• D. None of these

Answer 1

The correct option is A ( a ) No. of $\alpha -$ particles$=8$, No. of $\beta -$ particles$=6$; ( b ) ${}_{82}P{b}^{206}$

i) ${}_{92}{U}^{238}\to {{}_{82}Pb}^{206}+n{{}_{2}He}^4+n{{}_{-1}e}^0$
Equating mass number of both sides $238=206+4n+n\times 0\Rightarrow n=8$
Equation atomic number on both sides $92=82+2n+n(-1)=82+2\times 8+8(-1)\Rightarrow n=6$
Number of $\alpha -$ particles $=8$
Number of $\beta -$ particles $=6$
ii) On balancing the mass number and charge of the nuclear reaction, we obtain
${}^{}{Th}^{234}\to 7_2{He}^4+6_{-1}{e}^0{+}_{82}{Pb}^{208}$
So stable particle is ${}^{}{Pb}^{208}$