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\(^{202}_{84} Y\) → \(^4_2\)He + \(^0_{-1}\)e + \(^{98}_{83}\)X Therefore, the atomic number of the resulting nuclide = 83.

Using N = N\(_o\)\((\frac{1}{2})^n\) N\(_o\) = 48g, N =1.5g and n = number of half-lives 1.5 = 48\((\frac{1}{2})^n\) \(\frac{1.5}{48}\) = \(\frac{1}{32}\) = (\(\frac{1}{2})^5\) = \((\frac{1}{2})^n\) n = 5 (same base, equate power) Total time = n x t\(_{\frac{1}{2}}\) = 5 x 14 = 70 days Therefore, 1.5g will remain after 70 days.

To calculate the energy released during the fission process, we can use Einstein's mass-energy equivalence principle, given by the formula: E = \Deltamc\(^2\) m = 1.0 x 10\(^{-2}\)kg c = 3.0 x 10\(^8\)ms\(^{-1}\) Mass decrease = 0.02% \Deltam = 0.02% x 1.0 x 10\(^{-2}\) = 2.0\times10\(^{−6}\)kg E = \Deltamc\(^2\) = 2.0\times10\(^{−6}\) x (3.0 x 10\(^8\))\(^2\) = 1.8\times10\(^{11}\)J

Absorption line spectra are produced when electrons in an atom absorb specific wavelengths of light to move from a lower energy level to a higher energy level (excitation). The absorbed energy corresponds to the difference in energy between the two levels, resulting in dark lines in the spectrum at those specific wavelengths.

To arrange the processes in the correct order for generating electricity in a nuclear reactor, let's break down the steps involved: 1. Heat Generation: The nuclear reaction generates heat. 2. Heat Removal: This heat is removed by passing water through the reactor. 3. Steam Production: The heated water then passes through a heat exchanger to produce steam. 4. Turbine Operation: The steam drives turbines. 5. Electricity Generation: The turbines generate electricity Therefore, the correct order is: ii: The heat energy released is removed by passing water through the reactor. iv: The water then passes through some form of heat exchanger to produce steam. i: The steam is used to drive turbines. iii: The turbines in turn generate electricity

The duality of matter refers to the concept in quantum mechanics that matter exhibits both wave-like and particle-like properties.

The largest quantum of energy a hydrogen atom liberates occurs during a transition from a higher to a lower energy level. The energy difference between levels can be calculated using the formula: E = −13.6eV(\(\frac{1}{n_f^2} - \frac{1}{n_i^2​}\)) n\(_i\)​ is the initial quantum number (higher energy level), n\(_f\) is the final quantum number (lower energy level. A. n = 2 to n = 1 ​

2E - E = \(\frac{hc}{\lambda}\) = E = \(\frac{hc}{\lambda}\) \(\frac{4}{3}\)E = \(\frac{hc}{\lambda_{new}}\) E = \(\frac{3hc}{\lambda_2}\) \(\frac{3hc}{\lambda_2}\) = \(\frac{hc}{\lambda}\) \(\lambda_{new}\) = 3 \(\lambda_o\)

Nucleon number = Proton number + Neutron number Neutron number = Nucleon number − Proton number Neutron number = 23 - 11 = 12 The number of neutrons present in the atom is 12

The e.m.f of the accumulator = 1.5v Table of value/observations R(\(\Omega\)) 1\(_{amp}\) 1\(^{-1}_{amp}\) 0 0.78 1.28 1 0.50 2.00 2 0.38 2.63 3 0.30 3.33 4 0.25 4.00 5 0.22 4.55 Slope (s) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{4-1}{4-2} = \frac{3}{2}\) The intercept on the vertical axis is =-1.85\(\Omega\) Precautions: I tightened connections. I avoided errors due to parallax in reading ammeter or voltmeter. I noted zero error of ammeter/ voltmeter. I removed key when readings were not taken. I took repeated readings from the table, i.e. averaging the reading. (b)i. - The lead-acid accumulator can be recharged while the Leclanche cell cannot. - The lead-acid accumulator has the capacity to supply large currents for a long time while the leclanche cell does not. ii. The two resistors 3\(\Omega\) and 4\(\Omega\) are connected in parallel. \(\therefore\) The equivalent resistance (\(\Omega\) = \(\frac{1}{R} = \frac{1}{3} = \frac{1}{4} = \frac{4+3}{12}\) = \(\frac{7}{12}\) = 1.71 The effective resistance in the circuit = 1.71 + 4 = 5.71\(\Omega\)

Weight of handbag(W) = 162N acting vertically downward Angle of pull (\(\theta\)) = 60º to the vertical Since the handbag is in equilibrium, the total vertical force must equal the weight of the handbag. Let F be the force exerted by each student on the handle. The vertical component of the force from each student is given by: Fv = F cos 60º but there are two students 2Fcos60º = 162 2F(0.5) = 162 ( cos60º = 0.5) F = 162 ( since 2 x 0.5 = 1) Therefore, the force exerted by each student on the handbag is: 162N ​

The formula for Force on the particle is Bqv.

Based on nuclear physics, the correct answer is B: \(^3_2\)He has two protons and one neutron.

The process by which free electrons are emitted from the surface of a metal when external heat energy is applied is called thermionic emission.

No expanded working has been attached for this question yet. The verified answer is Option C.