Explanation
Different values of magnetic azimuthal quantum number are m1 = 2l + 1 = 2(3) + 1 = 7.
A represents series limit of Lyman series, B represents third member of Balmer series and C represents second member of Paschen series.
Total energy of electron in excited state = -13.6 + 12.1 = -1.5 eV, which corresponds to third orbit. The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st.
Increase of charge no. by 1 indicates β emission. Decrease of mass number by 4 and charge no. by 2 indicates α emission. No change of mass no. and charge no. indicates γ-emission
Average BE/nucleon increases first, and then decreases, as is clear from BE curve.
An electron is accompanied by an antineutrino.
In fusion, two lighter nuclei combine to give a heavier nucleus and possibly other products.
This is the basis of Brehmstrahlung radiations.
This is because neutrons are neutral. They carry no charge. They are, therefore, unaffected by electric fields.
Neutrons are fermions. They obey Fermi Dirac statistics.
Nuclear force of attraction between any two nucleons (n – n, p – p ; p – n) is same. The difference comes up only due to electrostatic force of repulsion between two protons.
F1 = F2 ≠ F2 . As F2 > F3 or F1
F1 = F3 > F2.
For γ – rays, λ ≈ 10-12 m
As disintegration ‘by two different processes is simultaneous, therefore, effective decay constant (λ = (λ1 + λ2).
A and C are isotopes as their charge no. is same
Applying principle of energy conservation, Energy of proton = total B.E. of 2α-energy of Li7 = 8 × 7.06 – 7 × 5.6 = 56.48 – 39.2 = 17.28 MeV
U-235 and P-239 undergo fission by thermal neutrons.
Lowest-orbit corresponds to n = 1. Energy is negative max i.e. the minimum.
Decrease in mass no. = 4 Decrease in charge no. = 2 – 1 = 1. ∴ (d) is correct.
Decrease in mass no. = 8 × 4 + 5 × 0 = 32 Decrease in charge no. = 8 × 2 – 5 × 1 = 11. Therefore (a) is the right choice.
Conservation of mass number and charge number show that both have zero value for X. Therefore, X must be neutrino only.
The nuclear reaction can be put as 6C11 → 5B11 + 1e0 + zXA Applying conservation of mass number and charge number, we find that A = 0 and Z = 0. Therefore, X stands for a neutrino.
The infrared radiations have lower energy than ultra violet radiations. Therefore, the only possible transition is from n = 5 to n = 4. As n increases, difference of energy levels and hence energy emitted decreases.
In β-decay, a neutron decays into a proton and an electron, which is emitted.
Isobars have same mass number and different charge number. β-emission involves change in charge number alone. Hence it produces isobars.
Total mass before fusion = 2m mass converted into energy in this process = E/c2. ∴ Mass of helium formed = (2m -E/c2).
Decrease in charge number due to 8α emissions = 8 × 2 = 16. Increase in charge number due to 4β− emissions = 4 × 1 = 4. Decrease in charge number due to 2β+ emission = 2 × 1 = 2. Net decrease in charge number = 16 – 4 + 2 = 14. ∴ Z of resulting nucleus = 92 – 14 = 78.
Protons cannot be emitted from radioactive nuclei.
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