🔑:The theory that atoms have a compact nucleus is supported by a wealth of evidence from various scientific experiments and observations. One of the most significant contributions to this understanding came from Ernest Rutherford's experiment with alpha particles and gold foil, which is often referred to as the "gold foil experiment" or "Rutherford's experiment." Here's a summary of the evidence and Rutherford's experiment:Background:In the early 20th century, the prevailing model of the atom was the "plum pudding" model, proposed by J.J. Thomson. According to this model, atoms were composed of a positively charged sphere (the "pudding") with negatively charged electrons (the "plums") embedded within. However, this model was soon challenged by Rutherford's experiment.Rutherford's Experiment (1909):Ernest Rutherford, a New Zealand-born physicist, designed an experiment to test the plum pudding model. He bombarded a thin layer of gold foil with alpha particles, which are high-energy helium nuclei. The alpha particles were emitted by a radioactive source and had a relatively large mass and positive charge. Rutherford expected the alpha particles to pass through the gold foil with minimal deflection, as predicted by the plum pudding model.Observations:To Rutherford's surprise, the alpha particles did not pass through the gold foil as expected. Instead, some of the particles were deflected at very large angles, and a small fraction were even deflected back towards the source. This was a shocking result, as it indicated that the alpha particles had encountered a dense, positively charged region within the atom.Interpretation:Rutherford's observations led him to propose the "nuclear atom" model, which posits that atoms have a small, dense, positively charged nucleus at their center. The nucleus is surrounded by a cloud of negatively charged electrons, which occupy a much larger volume. The alpha particles that were deflected at large angles had interacted with the nucleus, while those that passed through the gold foil with minimal deflection had interacted with the electrons.Key Evidence:Rutherford's experiment provided several key pieces of evidence that support the theory of a compact nucleus:1. Large-angle deflections: The deflection of alpha particles at large angles indicated that the nucleus is small and dense, with a positive charge.2. Backscattering: The observation of alpha particles being deflected back towards the source suggested that the nucleus is a compact, positively charged region.3. Scattering patterns: The distribution of scattered alpha particles revealed that the nucleus is much smaller than the atom as a whole.Additional Evidence:Other experiments and observations have since confirmed the existence of a compact nucleus:1. X-ray scattering: Experiments with X-ray scattering have shown that the nucleus is responsible for the scattering of X-rays, which is consistent with the nuclear atom model.2. Neutron scattering: Neutron scattering experiments have also provided evidence for the existence of a compact nucleus.3. Particle accelerators: The study of high-energy particle collisions has confirmed the existence of a small, dense nucleus at the center of atoms.In conclusion, Rutherford's experiment with alpha particles and gold foil provided crucial evidence for the existence of a compact nucleus at the center of atoms. The observations of large-angle deflections, backscattering, and scattering patterns all supported the nuclear atom model, which has since been confirmed by numerous other experiments and observations.

🔑:## Step 1: Calculate the volume of the spherical shellTo find the volume of the spherical shell, we use the formula for the volume of a sphere, (V = frac{4}{3}pi r^3), and then subtract the volume of the inner sphere from the volume of the outer sphere. The volume of the shell is given by (V_{shell} = frac{4}{3}pi (R_{outer}^3 - R_{inner}^3)), where (R_{outer} = 15.0) cm and (R_{inner} = 14.6) cm.## Step 2: Convert the radii from cm to mTo use the values in calculations involving SI units, we convert the radii from centimeters to meters: (R_{outer} = 0.15) m and (R_{inner} = 0.146) m.## Step 3: Calculate the volume of the shell in cubic metersSubstitute the values of (R_{outer}) and (R_{inner}) into the formula for the volume of the shell: (V_{shell} = frac{4}{3}pi (0.15^3 - 0.146^3)).## Step 4: Perform the calculation for the volume of the shell(V_{shell} = frac{4}{3}pi (0.15^3 - 0.146^3) = frac{4}{3}pi (0.003375 - 0.003089) = frac{4}{3}pi (0.000286) approx 0.0001201) m(^3).## Step 5: Calculate the charge density of the shellThe charge density (rho) is given by the total charge (Q) divided by the volume of the shell, (V_{shell}): (rho = frac{Q}{V_{shell}}). Given (Q = 5.85 times 10^{-8}) C, we calculate (rho).## Step 6: Perform the calculation for charge density(rho = frac{5.85 times 10^{-8}}{0.0001201} approx 487.31) C/m(^3).## Step 7: Apply Gauss's Law to find the electric field at the center, inner radius, and outer radiusFor a spherical shell, the electric field inside the shell (including at the center) is zero. At the surface and outside, the electric field can be found using (E = frac{kQ}{r^2}), where (k) is Coulomb's constant ((k approx 8.99 times 10^9) N(cdot)m(^2)/C(^2)) and (r) is the distance from the center of the shell.## Step 8: Calculate the electric potential at the center of the shellThe electric potential (V) at a point due to a point charge or a spherical charge distribution is given by (V = frac{kQ}{r}). However, for points inside a uniformly charged spherical shell, the potential is constant and equal to the potential at the surface, since the electric field inside is zero.## Step 9: Calculate the electric potential at the inner and outer radiiThe potential at the inner radius ((r = R_{inner})) and the outer radius ((r = R_{outer})) will be the same as the potential at the surface, given by (V = frac{kQ}{R_{outer}}), since the charge is distributed uniformly throughout the volume of the shell.## Step 10: Perform the calculation for electric potentialGiven (Q = 5.85 times 10^{-8}) C and (R_{outer} = 0.15) m, (V = frac{8.99 times 10^9 times 5.85 times 10^{-8}}{0.15} approx 348.41) V.The final answer is: boxed{348.41}

🔑:## Step 1: Define the problem and the systemWe have a 2D square lattice of point masses connected by springs. Each point mass can only move vertically, and we are considering the Lagrangian for this system. The Lagrangian is defined as the difference between the kinetic energy (T) and the potential energy (U) of the system: L = T - U.## Step 2: Determine the kinetic energy of the systemThe kinetic energy (T) of the system can be calculated by summing the kinetic energies of each point mass. For a point mass m at position (x, y) with vertical displacement from equilibrium q(x, y), its kinetic energy is given by (T = frac{1}{2}mdot{q}(x, y)^2), where (dot{q}(x, y)) is the vertical velocity of the mass. For a lattice with N masses, (T = sum_{i=1}^{N} frac{1}{2}mdot{q}_i^2), where (q_i) represents the vertical displacement of the i-th mass.## Step 3: Determine the potential energy of the systemThe potential energy (U) of the system comes from the springs connecting the point masses. For a spring with spring constant k and equilibrium length l, the potential energy due to a displacement (Delta l) from equilibrium is given by (U = frac{1}{2}k(Delta l)^2). In a 2D lattice, each internal point mass is connected to four neighbors (up, down, left, right), and edge masses are connected to three neighbors, while corner masses are connected to two neighbors. The total potential energy is the sum of the potential energies of all springs.## Step 4: Consider non-linear interactionsNon-linear interactions can arise from considering higher-order terms in the expansion of the potential energy of the springs. The simplest non-linear interaction would be the cubic term in the expansion of the spring potential energy, representing the anharmonicity of the spring. However, in the context of deriving the basic Lagrangian for a 2D square lattice of point masses connected by springs, we typically first consider the harmonic (linear) approximation for simplicity.## Step 5: Formulate the LagrangianGiven the kinetic energy (T = sum_{i=1}^{N} frac{1}{2}mdot{q}_i^2) and the potential energy (U = sum_{text{springs}} frac{1}{2}k(Delta l)^2), the Lagrangian L is (L = T - U = sum_{i=1}^{N} frac{1}{2}mdot{q}_i^2 - sum_{text{springs}} frac{1}{2}k(Delta l)^2). For a 2D lattice with only vertical displacements, (Delta l) can be expressed in terms of the displacements (q_i) of the masses.## Step 6: Discuss physical implicationsThe potential energy terms in the Lagrangian imply that the system's energy increases as the masses are displaced from their equilibrium positions, due to the stretching or compressing of springs. Non-linear interactions, if considered, would introduce terms that depend on higher powers of the displacements, leading to more complex dynamics, such as anharmonic oscillations.The final answer is: boxed{L = sum_{i=1}^{N} frac{1}{2}mdot{q}_i^2 - sum_{text{springs}} frac{1}{2}k(Delta l)^2}

🔑:## Step 1: Recall Snell's LawSnell's Law states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities of the two media, or equivalently, the inverse ratio of the refractive indices. The equation is (n_1 sin(theta_1) = n_2 sin(theta_2)), where (n_1) and (n_2) are the refractive indices of the first and second media, respectively, and (theta_1) and (theta_2) are the angles of incidence and refraction, respectively.## Step 2: Apply Snell's Law to Total ReflectionFor total reflection to occur, the angle of refraction ((theta_2)) must be 90 degrees, which means the light is completely reflected back into the first medium and does not pass into the second medium. The sine of 90 degrees is 1. In our case, (n_1 = 1.4) (glass) and (n_2 = 1) (air). We are looking for the critical angle ((theta_c)), which is the angle of incidence ((theta_1)) when (theta_2 = 90^circ). So, we set (theta_2 = 90^circ) in Snell's Law.## Step 3: Solve for the Critical AnglePlugging the values into Snell's Law, we get (1.4 sin(theta_c) = 1 sin(90^circ)). Since (sin(90^circ) = 1), the equation simplifies to (1.4 sin(theta_c) = 1). To solve for (theta_c), we divide both sides by 1.4, resulting in (sin(theta_c) = frac{1}{1.4}).## Step 4: Calculate the Sine InverseTo find (theta_c), we take the sine inverse (or arcsine) of (frac{1}{1.4}). So, (theta_c = arcsinleft(frac{1}{1.4}right)).## Step 5: Compute the ValueUsing a calculator, (arcsinleft(frac{1}{1.4}right) approx arcsin(0.7142857)).The final answer is: boxed{45.58}