Atomic Structure and Hydrogen Atom

<p>Thomson's model/ Rutherford's model</p> Signup and view all the answers

<p>Rutherford's model</p> Signup and view all the answers

<p>Since hydrogen has a much lighter nucleus compared to gold, the alpha particles would be deflected by much smaller angles, meaning that most of the alpha particles would pass through the hydrogen foil without significant deflection. The number of particles scattered backwards would be very low.</p> Signup and view all the answers

<p>The frequency of the emitted radiation can be calculated using the relationship: hv = E_i - E_f, where h is Planck's constant (6.63 × 10^-34 Js), v is the frequency, and E_i and E_f are the energies of the initial and final states, respectively. Converting the energy difference of 2.3 eV to joules (1 eV = 1.602 × 10^-19 J) and plugging the values into the equation, we get: v = (2.3 × 1.602 × 10^-19 J) / (6.63 × 10^-34 Js) = 5.56 × 10^14 Hz.</p> Signup and view all the answers

<p>The total energy of the electron in the ground state is -13.6 eV. This energy is the sum of its kinetic energy (K) and potential energy (U). The kinetic energy of the electron is half its potential energy, so K = -13.6 eV / 2 = -6.8 eV. The potential energy is then U = -13.6 eV - (-6.8 eV) = -6.8 eV.</p> Signup and view all the answers

<p>The energy difference between the n=4 and n=1 levels in a hydrogen atom is 12.75 eV. Therefore, the photon absorbed by the atom has an energy of 12.75 eV. Using the relationship E = hc/λ, where h is Planck's constant (6.63 × 10^-34 Js), c is the speed of light (3 × 10^8 m/s), and λ is the wavelength, we can calculate the wavelength of the photon: λ = hc/E = (6.63 × 10^-34 Js)(3 × 10^8 m/s) / (12.75 × 1.602 × 10^-19 J) = 9.72 × 10^-8 m = 97.2 nm. The frequency of the photon can be calculated using the equation ν = c/λ = (3 × 10^8 m/s) / (9.72 × 10^-8 m) = 3.09 × 10^15 Hz.</p> Signup and view all the answers

<p>The speed of the electron in the nth Bohr orbit is given by v = (e^2)/(2ε_0hn) = (1.602 × 10^-19 C)^2 / (2 × 8.85 × 10^-12 C^2/(Nm^2) × 6.63 × 10^-34 Js × n). Substituting the values for n = 1, 2, and 3, we get: v_1 = 2.19 × 10^6 m/s, v_2 = 1.09 × 10^6 m/s, and v_3 = 7.3 × 10^5 m/s. The orbital period (T) in the nth Bohr orbit is given by T = (2πr_n)/v, where r_n is the radius of the nth Bohr orbit. Substituting the value of r_n = n^2 a_0, where a_0 is the Bohr radius (5.29 × 10^-11 m), we get: T_1 = (2π) × (5.29 × 10^-11 m) / (2.19 × 10^6 m/s) = 1.52 × 10^-16 s, T_2 = (2π) × (4 × 5.29 × 10^-11 m) / (1.09 × 10^6 m/s) = 1.22 × 10^-15 s, and T_3 = (2π) × (9 × 5.29 × 10^-11 m) / (7.3 × 10^5 m/s) = 3.88 × 10^-15 s.</p> Signup and view all the answers

<p>The radii of the orbits in the Bohr model are quantized and directly proportional to the square of the principal quantum number (n). Therefore, the radius of the n = 2 orbit is 2^2 * 5.3 × 10^-11 m = 2.12 × 10^-10 m, and the radius of the n = 3 orbit is 3^2 * 5.3 × 10^-11 m = 4.8 × 10^-10 m.</p> Signup and view all the answers

<p>When a hydrogen atom absorbs energy from an electron beam, its electron can be excited to higher energy states. The electron will eventually transition back to lower energy states, emitting photons with energies equal to the energy differences between these states. The 12.5 eV electron beam has enough energy to excite the electron to the n = 2 or higher levels. Therefore, the emitted photons will correspond to transitions from these higher levels to the n = 1 level, resulting in the emission of the Lyman series, which is characterized by ultraviolet wavelengths.</p> Signup and view all the answers

<p>To find the quantum number (n) that characterizes Earth's orbit around the Sun using Bohr's model, we can apply the quantization condition for angular momentum: mvr = nh/2π. Given the radius (r) of the orbit is 1.5 × 10^11 m, the speed (v) of Earth is 3 × 10^4 m/s, and the mass (m) of Earth is 6.0 × 10^24 kg, Planck's constant (h) is 6.63 × 10^-34 Js. Plugging these values into the equation yields: n = (2πmvr) / h = (2π × 6.0 × 10^24 kg × 3 × 10^4 m/s × 1.5 × 10^11 m) / (6.63 × 10^-34 Js) ≈ 2.55 × 10^74. This extremely large value of ‘n’ highlights that Bohr's model is not appropriate for macroscopic systems like the solar system. It was designed to model the behavior of electrons within atoms.</p> Signup and view all the answers