Question
Question: A\[Li\] target is bombarded with a proton beam current of\[{10^{ - 4}}\]\[A\] for one hour to produc...
ALi target is bombarded with a proton beam current of{10^{ - 4}}$$$$A for one hour to produce Be of activity 1.8×108dps. Assuming that one Be radioactive nuclei is produced by bombarding 1000 protons, its half-life is:
(a) 0.87 \times {10^7}$$$$s
(b) 0.2 \times {10^7}$$$$s
(c) 0.67 \times {10^8}$$$$s
(d) 0.87 \times {10^6}$$$$s
Solution
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 (I)=10−4A, Time(t)= 1h=3600s
We also know that
Q=It
Where, Qis electric charge, I is electric current and t is time.
Q=10−4×1×3600=0.36C
Number of protons = eQ = 1.6×10−190.36
Number of protons =2.25×108
Produced number of Be nuclei = 10002.25×1018
= 2.25×1015
Given that, activity of Be = 1.8×108dps
∴ Activity =λN
Activity= t1/2ln2N
∴ λ=t1/2ln2
Given that, value of activity = 1.8×108dps
1.8×108=t1/20.693×2.25×1015
Then,
t1/2=1.8×1080.693×2.25×1015
t1/2=0.87×107s
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.