
A\[Li\] target is bombarded with a proton beam current of\[{10^{ - 4}}\]\[A\] for one hour to produce \[Be\] of activity \[1.8 \times {10^8}dps\]. Assuming that one \[Be\] radioactive nuclei is produced by bombarding \[1000\] protons, its half-life is:
(a) \[0.87 \times {10^7}\]\[s\]
(b) \[0.2 \times {10^7}\]\[s\]
(c) \[0.67 \times {10^8}\]\[s\]
(d) \[0.87 \times {10^6}\]\[s\]
Answer
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Hint: Atomic number – Atomic number is the number of protons in the nucleus of an atom.
Atomic number of lithium (\[Li\]) =\[3\]
Atomic number of beryllium (\[Be\]) = 4
A nucleus consists of protons, electrons and neutrons.
Protons – protons are a type of subatomic particle with positive charge.
Electrons – Electrons are a type of subatomic particles with negative charge.
Neutrons – Neutrons are subatomic particles with no charge.
Complete step by step answer:
Given that, current \[\left( I \right)\]=\[{10^{ - 4}}A\], Time\[\left( t \right)\]= \[1h\]=\[3600s\]
We also know that
\[Q = It\]
Where, \[Q\]is electric charge, \[I\] is electric current and \[t\] is time.
\[Q = {10^{ - 4}} \times 1 \times 3600\]=\[0.36C\]
Number of protons = \[\dfrac{Q}{e}\] = \[\dfrac{{0.36}}{{1.6 \times {{10}^{ - 19}}}}\]
Number of protons =\[2.25 \times {10^8}\]
Produced number of \[Be\] nuclei = \[\dfrac{{2.25 \times {{10}^{18}}}}{{1000}}\]
= \[2.25 \times {10^{15}}\]
Given that, activity of \[Be\] = \[1.8 \times {10^8}dps\]
∴ Activity =\[\lambda N\]
Activity= \[\dfrac{{\ln 2}}{{{t_{1/2}}}}N\]
∴ \[\lambda = \dfrac{{\ln 2}}{{{t_{1/2}}}}\]
Given that, value of activity = \[1.8 \times {10^8}dps\]
\[1.8 \times {10^8} = \dfrac{{0.693}}{{{t_{1/2}}}} \times 2.25 \times {10^{15}}\]
Then,
\[{t_{1/2}} = \dfrac{{0.693 \times 2.25 \times {{10}^{15}}}}{{1.8 \times {{10}^8}}}\]
\[{t_{1/2}} = 0.87 \times {10^7}s\]
So, the correct answer is “Option A”.
Note:
Radioactive decay – Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiations. When an unstable atom gets radioactive then the number of protons change in the nucleus. Radioactive sources are used to treat diseases, to produce energy, to sterilize instruments. Sometimes nucleus breaks, undergoing nuclear decay. All elements with 84 or more protons are unstable and undergo decay.
Atomic number of lithium (\[Li\]) =\[3\]
Atomic number of beryllium (\[Be\]) = 4
A nucleus consists of protons, electrons and neutrons.
Protons – protons are a type of subatomic particle with positive charge.
Electrons – Electrons are a type of subatomic particles with negative charge.
Neutrons – Neutrons are subatomic particles with no charge.
Complete step by step answer:
Given that, current \[\left( I \right)\]=\[{10^{ - 4}}A\], Time\[\left( t \right)\]= \[1h\]=\[3600s\]
We also know that
\[Q = It\]
Where, \[Q\]is electric charge, \[I\] is electric current and \[t\] is time.
\[Q = {10^{ - 4}} \times 1 \times 3600\]=\[0.36C\]
Number of protons = \[\dfrac{Q}{e}\] = \[\dfrac{{0.36}}{{1.6 \times {{10}^{ - 19}}}}\]
Number of protons =\[2.25 \times {10^8}\]
Produced number of \[Be\] nuclei = \[\dfrac{{2.25 \times {{10}^{18}}}}{{1000}}\]
= \[2.25 \times {10^{15}}\]
Given that, activity of \[Be\] = \[1.8 \times {10^8}dps\]
∴ Activity =\[\lambda N\]
Activity= \[\dfrac{{\ln 2}}{{{t_{1/2}}}}N\]
∴ \[\lambda = \dfrac{{\ln 2}}{{{t_{1/2}}}}\]
Given that, value of activity = \[1.8 \times {10^8}dps\]
\[1.8 \times {10^8} = \dfrac{{0.693}}{{{t_{1/2}}}} \times 2.25 \times {10^{15}}\]
Then,
\[{t_{1/2}} = \dfrac{{0.693 \times 2.25 \times {{10}^{15}}}}{{1.8 \times {{10}^8}}}\]
\[{t_{1/2}} = 0.87 \times {10^7}s\]
So, the correct answer is “Option A”.
Note:
Radioactive decay – Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiations. When an unstable atom gets radioactive then the number of protons change in the nucleus. Radioactive sources are used to treat diseases, to produce energy, to sterilize instruments. Sometimes nucleus breaks, undergoing nuclear decay. All elements with 84 or more protons are unstable and undergo decay.
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