As one moves along the line of stability from 56Fe to 235U nucleus, the nuclear binding energy per particle decreases from about 8.8 MeV to 7.6 MeV. This trend is mainly due to the
A. short range nature of the nuclear forces
B. long range nature of the Coulomb forces
C. tensor nature of the nuclear forces
D. spin dependence of the nuclear forces
Answer: Option B
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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