Let $$\left| n \right\rangle $$ and $$\left| p \right\rangle $$ denote the isospin state with $$I = \frac{1}{2},\,{I_3} = \frac{1}{2}$$ and $$I = \frac{1}{2},\,{I_3} = - \frac{1}{2}$$ of a nucleon respectively. Which one of the following two-nucleon states has $$I = 0,\,{I_3} = 0\,?$$
A. $$\frac{1}{{\sqrt 2 }}\left( {\left| {nn} \right\rangle - \left| {pp} \right\rangle } \right)$$
B. $$\frac{1}{{\sqrt 2 }}\left( {\left| {nn} \right\rangle + \left| {pp} \right\rangle } \right)$$
C. $$\frac{1}{{\sqrt 2 }}\left( {\left| {np} \right\rangle - \left| {pn} \right\rangle } \right)$$
D. $$\frac{1}{{\sqrt 2 }}\left( {\left| {np} \right\rangle + \left| {pn} \right\rangle } \right)$$
Answer: Option C
A. Thorium series
B. Neptunium series
C. Uranium series
D. Actinium series
A. 10-10 eV
B. 10-9 eV
C. 10-6 eV
D. 10-4 eV
A. The process is allowed because ΔS = 0
B. The process is allowed because $$\Delta {I_3} = 0$$
C. The process is not allowed because ΔS ≠ 1 and $$\Delta {I_3} \ne 0$$
D. The process is not allowed because the Baryon number is violated
A. $${\left( {{}^1{s_{1/2}}} \right)^2}{\left( {{}^1{p_{3/2}}} \right)^3};\,J = \frac{3}{2}$$
B. $${\left( {{}^1{s_{1/2}}} \right)^2}{\left( {{}^1{p_{1/2}}} \right)^2}{\left( {{}^1{p_{3/2}}} \right)^1};\,J = \frac{3}{2}$$
C. $${\left( {{}^1{s_{1/2}}} \right)^1}{\left( {{}^1{p_{3/2}}} \right)^4};\,J = \frac{1}{2}$$
D. $${\left( {{}^1{s_{1/2}}} \right)^2}{\left( {{}^1{p_{3/2}}} \right)^2}{\left( {{}^1{p_{1/2}}} \right)^1};\,J = \frac{1}{2}$$
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