Question

A $\_{}^{7}Li$ target is bombarded with a proton beam current of magnitude $10^{- 4} \, A$ for an hour to produce $\_{}^{7}Be$ of activity $1.8\times 10^{8} \, $ disintegrations per second. If we assume that one radioactive nucleus of beryllium is produced by bombarding $1000$ protons, then its half-life is close to

• A. $100$ days
• B. $110$ days
• C. $150$ days
• D. $80$ days

Answer 1

Answer: (A)

At time $t$ , let say there are $N$ atoms of $\_{}^{7}Be$ (radioactive). Then the net rate of formation of $\_{}^{7}Be$ nuclei at this instant is,
$\frac{d N ⁡}{d ⁡ t} = \frac{1 0^{- 4}}{1 \text{.} 6 \times 1 0^{- 1 9} \times 1 0 0 0} - \lambda N ⁡$
or $\frac{d N ⁡}{d ⁡ t} = 6 \text{.} 2 5 \times 1 0^{1 1} - \lambda N ⁡$
or $\displaystyle \int _{o }^{N ⁡_{0}} \frac{d ⁡ N ⁡}{6 \text{.} 2 5 \times 1 0^{1 1} - \lambda N ⁡} = \displaystyle \int _{0}^{3 6 0 0} d ⁡ t$
where $N_{0}$ are the number of nuclei at $t=1hour$ or $3600s$ .
∴ $- \frac{1}{\lambda } \text{ln} \left(\frac{6 \text{.} 2 5 \times 1 0^{1 1} - \lambda \left(N \right)_{0}}{6 \text{.} 2 5 \times 1 0^{1 1}}\right) = 3 6 0 0$
$\lambda N_{0}$ = activity of $\_{}^{7}Be$ at $t=1hour$ $=1.8\times 10^{8}$ disintegrations per second
∴ $- \frac{1}{\lambda } \text{ln} \left(\frac{6 \text{.} 2 5 \times 1 0^{1 1} - 1 \text{.} 8 \times 1 0^{8}}{6 \text{.} 2 5 \times 1 0^{1 1}}\right) = 3 6 0 0$
∴ $\lambda =8.0\times 10^{- 8}s^{- 1}$
Therefore, half-life $t_{1 / 2}=\frac{0 \text{.} 6 9 3}{8 \text{.} 0 \times 1 0^{- 8}}=8\text{.}66\times 10^{6}s$
$=100.26$ days