Question

U with atomic number 92 and mass number 238 is radioactive and it emits alpha and beta particles to form Pb with atomic number 82 and mass number 206. Calculate the number of alpha and beta particles emitted in this conversation.

An ore of U with atomic number 92 and mass number 238 is found to contain U and Pb in the weight ratio of 1:0.1. The half-life period of U is $4.5\times {10}^9$ years. Calculate the age of the ore.

• A. $6.7/times{10}^9$
• B. $7.12/times{10}^8$
• C. $8.5/times{10}^{10}$
• D. $6.6/times{10}^9$

Answer 1

Answer: (B)

(i) The equation for the nuclear reaction is given by,

${}_{92}{U}^{238}⟶{}_{82}{Pb}^{206}+x_2{He}^4+y_{-1}{e}^0$

where $x$ = No. of alpha particles emitted in the conversion

$y$ = No. of beta particles emitted in the conversion

$92=82+2x-y\Rightarrow 2x-y=10$

$238=206+4x+0\Rightarrow x=8$

$\therefore y=6$

Hence the number of alpha and beta particles emitted in the conversation are 8 and 6 respectively.

(ii)

${}_{92}{U}^{238}⟶{}_{82}P{b}^{206}$

Time/Atom	U	Pb
t = 0	a	-
t = t	a-x	x

Here, $\frac{a-x}{x}=\frac{1}{0.1}\Rightarrow a=11x$

We know, $k=\frac{ln\left(2\right)}{t_{1/2}}=\frac{0.693}{4.5\times {10}^9}$

Now,

$t=\frac{2.303}{k}log\left(\frac{a}{a-x}\right)$

$t=\frac{2.303\times 4.5\times {10}^9}{0.693}log\left(\frac{11x}{11x-x}\right)=6.2\times {10}^8$ years.