Question

The half-life of $_{92}^{238}U$ undergoing $\alpha -$ decay is $4.5 \times 10^{9}$ $years$ . Calculate the activity of $1g$ sample of $_{92}^{238}U$ . (Avogadro's number is $6.022\times 10^{23}$ )

• A. $4.231\times 10^{4}dps$
• B. $1.721\times 10^{3}dps$
• C. $1.235\times 10^{4}dps$
• D. $5.167\times 10^{3}dps$

Answer 1

Answer: (C)

Given, $T_{1 / 2}=4.5\times 10^{9}years=4.5\times 10^{9}\times 3.156\times 10^{7} \, s$
Number of atoms in $1g$ Uranium,
$N = \frac{6.023 \times 10^{23}}{238}$ atoms
The activity of the sample is $\lambda \, N$
$=\frac{0.693}{T_{\frac{1}{2}}}\times N$
$=\frac{0.693 \times 6.023 \times 10^{23}}{4.5 \times 3.156 \times 10^{16} \times 238}$
$=1.235\times 10^{4} \, \text{ dps}.$