Question

The Half life of ${}_{92}^{238}U$ undergoing $\alpha$ decay is $4.5 \times 10^{9}$ years. The activity of 1 g sample of ${}_{92}^{238}U$ is

• A. $1.23 \times 10^{4}Bq$
• B. $1.23 \times 10^{5}Bq$
• C. $1.23 \times 10^{3}Bq$
• D. $1.23 \times 10^{6}Bq$

Answer 1

Answer: (A)

$1.23 \times 10^{4}Bq$

Answer 2

:$T_{1/2} = 4.5 \times 10^{9}y$

$= 4.5 \times 10^{9} \times 3.16 \times 10^{7} = 1.42 \times 10^{17}s$

One kmol of any isotope contains Avogadro’ number of atoms, so 1 g of ${}_{92}^{283}U$contains

$\frac{1}{2.38 \times 10^{- 3}}kmol \times 6.025 \times 10^{26}$atoms per kmol

$= 25.3 \times 10^{20}$atoms

Activity ,

$R = \lambda N = \frac{0.693}{T_{1/2}}N = \frac{0.693 \times 25.3 \times 10^{20}}{1.42 \times 10^{17}}s^{- 1}$

$= 1.23 \times 10^{4}s^{- 1} = 1.23 \times 10^{4}Bq$