A⁷ Li target is bombarded with a proton beam current of 10^-4 A for one hour to produce ⁷Be of activity 1.8 × 10⁸ disintegrations per second. Assuming that one ⁷Be radioactive nuclei is produced by bombarding 1000 protons, its half-life is

0.87 × 10⁷ s

0.2 × 10⁷ s

0.67 × 10⁸ s

0.87 × 10⁶ s

Solution

0.87 × 10⁷ s

Given:-

i = 10^-4 A

t = 1h = 3600 s

We know that

q = it

= 10^-4 × 1 × 3600

q = 0.36 C

∴ Number of protons = $\frac{q}{e}$

= $\frac{0.36}{1.6 \times 10^{- 19}}$ [ e = 1.6 × 10^-19 C ]

= 2.25 × 10⁸

Number of ⁷Be nuclei produced

= $\frac{2.25 \times 10^{8}}{1000}$

= 2.25 × 10⁸

Activity of ⁷Be = 1.8 × 10⁸ disintegration/s

Acivity A = λ N

= $\frac{\ln 2}{t_{\frac{1}{2}}}$ N $[∵ λ = \frac{\ln 2}{t_{\frac{1}{2}}}]$

1.8 × 10⁶ = $\frac{0.693}{t_{\frac{1}{2}}} \times 2.25 \times 10^{15}$

t_1/2 = $\frac{0.693 \times 2.25 \times 10^{15}}{1.8 \times 10^{8}}$

t_1/2 = 0.87 × 10⁷ s

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