JEE Physics Nuclei MCQs Set 31

Answer: (b) the high temperature maintained inside the reactor core

Explanation:
1. At the very high temperature the thermal energy exceeds the ionisation energy of the atoms.
2. Electrons are stripped from the nuclei, giving a fully ionised gas (a plasma).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) Deuteron density \( 8.0\times10^{14} \) cm\(^{-3}\), confinement time \( 9.0\times10^{-1} \) s

Explanation:
1. Lawson numbers \( nt_0 \): \( 1\times10^{10} \), \( 7.2\times10^{14} \), \( 4\times10^{12} \), \( 4\times10^{12} \).
2. Only the second design (\( 7.2\times10^{14}\gt 5\times10^{14} \)) exceeds the criterion.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) (iii) and (iv)

Explanation:
1. Ca-40 has a larger binding energy per nucleon than Ne-20, so its mass defect is more than twice that of Ne-20: \( M_2\lt 2M_1 \) (iii).
2. Every nucleus has mass less than the sum of its free nucleons: \( M_1\lt 10(m_n+m_p) \) (iv).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) (i) and (iv)

Explanation:
1. A stable nucleus has a positive binding energy, so its mass is less than that of the separate nucleons (i).
2. Fission releases energy by splitting a very heavy nucleus into medium-mass fragments with higher B/A (iv); fusing medium-mass nuclei does not release energy (iii false).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) (ii) and (iii)

Explanation:
1. Energy released from masses: \( \text C^{12}+\text H^1\rightarrow\text N^{13} \) gives \( (13.007825-13.005739)\times931.5=1.9 \) MeV (about 2); \( \text N^{14}+\text H^1\rightarrow\text O^{15} \) gives \( 0.007834\times931.5=7.3 \) MeV.
2. \( \text C^{13}+\text H^1\rightarrow\text C^{14} \) gives about 7.5 MeV, not 4.3. In (iv) the charge is \( 54+36=90\neq92 \), so it is not balanced.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) 24 MeV

Explanation:
1. \( \Delta m=2\times2.0141-4.0024=0.0258 \) u.
2. \( Q=0.0258\times931.5=24 \) MeV.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) 9

Explanation:
1. Energy available for the ground state: \( (12.014-12.000)\times931.5=13.04 \) MeV.
2. Decay to the excited state leaves \( 13.04-4.04=9.0 \) MeV for the \( \beta \)-particle (neglecting recoil).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) 3

Explanation:
1. The plant's power falls as \( P_0\left(\tfrac12\right)^{t/T} \) and must stay above 12.5% of \( P_0 \).
2. \( 12.5\%=\tfrac18=\left(\tfrac12\right)^3 \), so \( t=3T \), n = 3.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) A = 232 and Z = 90

Explanation:
1. Mass number: \( A=228+4=232 \).
2. Charge: \( Z=92-2=90 \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) 121 g

Explanation:
1. Thermal power \( =\dfrac{200\text{ MW}}{0.25}=8\times10^8 \) W, so the energy per day is \( 6.9\times10^{13} \) J. Energy per fusion \( =(2\times2.0141-4.0026)\times931.5=23.85 \) MeV \( =3.82\times10^{-12} \) J.
2. Fusions per day \( =1.81\times10^{25} \); deuterons \( =3.62\times10^{25} \); mass \( =3.62\times10^{25}\times2.0141\times1.66\times10^{-27}=0.121 \) kg \( =121 \) g.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) 6.3 MeV

Explanation:
1. \( p=\dfrac h\lambda=\dfrac{6.63\times10^{-34}}{5.76\times10^{-15}}=1.151\times10^{-19} \) kg m/s, so \( K_\alpha=\dfrac{p^2}{2m_\alpha}=6.2 \) MeV.
2. The daughter shares the same momentum, so its kinetic energy is \( K_\alpha\times\dfrac{4.002}{223.610}=0.11 \) MeV. Total \( \approx6.3 \) MeV.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) \( 6.25\times10^{11} \)

Explanation:
1. Incident power \( =2.0\times10^{-4} \) W; photons per second \( =\dfrac{2.0\times10^{-4}}{10.6\times1.6\times10^{-19}}=1.18\times10^{14} \).
2. Photoelectrons per second \( =0.0053\times1.18\times10^{14}=6.25\times10^{11} \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) 2 eV

Explanation:
1. The shorter wavelength is the 4 to 3 line: \( 13.6\times9\left(\tfrac19-\tfrac1{16}\right)=5.95 \) eV (the 5 to 4 line has only 2.75 eV).
2. \( \varphi=5.95-3.95=2 \) eV.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) 2

Explanation:
1. Six different photon energies means 4 levels are involved: \( \dfrac{4\times3}2=6 \).
2. After absorption the highest level is \( n=4 \). The lowest populated excited level is therefore \( n=4-2=2 \) (A is n = 1, B is n = 2, the absorbed photon takes B to n = 4).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) 14.4 eV

Explanation:
1. \( E_1\left(\tfrac14-\tfrac1{16}\right)=2.7\Rightarrow E_1\times\dfrac3{16}=2.7 \).
2. \( E_1=14.4 \) eV is the ionisation energy.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) 0, 11 and 13

Explanation:
1. A nucleus contains no electrons. Protons \( =Z=11 \).
2. Neutrons \( =A-Z=24-11=13 \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) 25

Explanation:
1. \( r_n=\dfrac{n^2a_0}{624} \).
2. \( r_n=a_0\Rightarrow n^2=624\Rightarrow n\approx25 \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.