Exams › JEE Advanced › Physics

Neon-23 undergoes beta-minus decay: Ne-23(10) -> Na-23(11) + beta⁻ + anti-neutrino. The atomic masses are: Ne-23 = 22.9945 u, Na-23 = 22.9898 u. Mass of electron = 0.51 MeV/c². Maximum kinetic energy of the beta⁻ particle (to nearest integer, in MeV). [Use 1 u = 931.5 MeV/c²]

1. 3
2. 4
3. 5
4. 6

Correct answer: 4

Solution

In beta-minus decay, using atomic masses, Q = [M_parent - M_daughter] * 931.5 MeV/u. The electron masses are already included in atomic masses and cancel. Q = (22.9945 - 22.9898) * 931.5 = 0.0047 * 931.5 ≈ 4.38 MeV ≈ 4 MeV.

Related JEE Advanced Physics questions

• The correct statement is:
• In the nuclear fission reaction 235⁹²U → 140⁵⁴Xe + 94³⁸Sr + x + y, where x and y represent two particles, the 235⁹²U nucleus is initially stationary. The kinetic energies of the resulting products are denoted as Kₓ, K_Sr, K_Xe (2 MeV), and K_y (2 MeV), respectively. The binding energy per nucleon values for 235⁹²U, 140⁵⁴Xe, and 94³⁸Sr are 7.5 MeV, 8.5 MeV, and 8.5 MeV, respectively. Considering the laws of conservation, which of the following is correct?
• An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?
• The electrostatic potential energy of Z protons uniformly distributed within a spherical nucleus of radius R is expressed as E = (3/5) * Z(Z−1)e² / (4πε₀R). The masses of neutron, ¹H, ¹⁵₇N, and ¹⁵₈O are 1.008665 u, 1.007825 u, 15.000109 u, and 15.003065 u, respectively. If the radii of ¹⁵₇N and ¹⁵₈O nuclei are identical, and given 1 u = 931.5 MeV/c², e²/(4πε₀) = 1.44 MeV fm, and 1 fm = 10⁻¹⁵ m, calculate the radius of either nucleus, assuming the binding energy difference between ¹⁵₇N and ¹⁵₈O arises solely from electrostatic energy.
• In a radioactive decay experiment, the starting number of nuclei is 3000. After 1.0 second, it is observed that 1000 ± 40 nuclei have undergone decay. Using the approximation ln(1 + x) ≈ x for |x| << 1, what is the uncertainty Δλ in calculating the decay constant λ (in s⁻¹)?
• A radioactive isotope 40/19 K undergoes decay either into stable 40/20 Ca with a decay constant of 4.5 × 10⁻¹⁰ per year or into stable 40/18 Ar with a decay constant of 0.5 × 10⁻¹⁰ per year. Assuming that all stable 40/20 Ca and 40/18 Ar nuclei originate from the decay of 40/19 K, determine the time t × 10⁹ years for which the ratio of the combined number of stable 40/20 Ca and 40/18 Ar nuclei to the remaining 40/19 K nuclei becomes 99. (Given ln 10 = 2.3).

⚔️ Practice JEE Advanced Physics free + battle 1v1 →