The radioactive isotope $$_{27}C{o^{60}}$$ which is used in the treatment of cancer can be made by $$(n, p)$$ reaction. For this reaction the target nucleus is
A.
$$_{28}N{i^{59}}$$
B.
$$_{27}C{o^{59}}$$
C.
$$_{28}N{i^{60}}$$
D.
$$_{27}C{o^{60}}$$
Answer :
$$_{28}N{i^{60}}$$
Solution :
$$_{28}N{i^{60}} + \mathop {_0{n^1}}\limits_{\left( n \right)} { \to _{27}}C{o^{60}} + \mathop {_1{H^1}}\limits_{\left( p \right)} $$
( $$n, p$$ means that neutron attacks and proton liberates )
Releted MCQ Question on Physical Chemistry >> Nuclear Chemistry
Releted Question 1
A nuclide of an alkaline earth metal undergoes radioactive decay by emission of three $$\alpha $$ - particles in succession. The group of the periodic table to which the resulting daughter element would belong is
The radioactive isotope $$_{27}C{o^{60}}$$ which is used in the treatment of cancer can be made by $$(n, p)$$ reaction. For this reaction the target nucleus is
The radio isotope, tritium $$\left( {_1{H^3}} \right)$$ has a half-life of $$12.3$$ $$yr.$$ If the initial amount of tritium is $$32$$ $$mg,$$ how many milligrams of it would remain after $$49.2$$ $$yr?$$