Atomic Structure and Hydrogen Atom

Podcast

Play an AI-generated podcast conversation about this lesson

Download our mobile app to listen on the go

Get App

Questions and Answers

The size of the atom in Thomson's model is ______ the atomic size in Rutherford's model.

much greater than (correct)
no different from
much less than

In the ground state of electrons are in stable equilibrium, while in electrons always experience a net force.

Rutherford's model/ Thomson's model
Thomson's model/ Rutherford's model (correct)

A classical atom based on ______ is doomed to collapse.

Rutherford's model (correct)
Thomson's model

An atom has a nearly continuous mass distribution in but has a highly non-uniform mass distribution in .

Thomson's model/ Rutherford's model (B) Signup and view all the answers

The positively charged part of the atom possesses most of the mass in ______.

Rutherford's model (A) Signup and view all the answers

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

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. Signup and view all the answers

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

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. Signup and view all the answers

The ground state energy of the hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?

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. Signup and view all the answers

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

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. Signup and view all the answers

Using the Bohr's model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. Calculate the orbital period in each of these levels.

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. Signup and view all the answers

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What are the radii of the n = 2 and n = 3 orbits?

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. Signup and view all the answers

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

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. Signup and view all the answers

In accordance with the Bohr's model, find the quantum number that characterizes the earth's revolution around the sun in an orbit of radius 1.5 × 10^11 m with orbital speed 3 × 10^4 m/s. (Mass of earth = 6.0 × 10^24 kg).

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. Signup and view all the answers

Flashcards

Atomic Structure

The arrangement of positive and negative charges within an atom.

Plum Pudding Model

An early model of the atom, proposing that positive charge is spread throughout the atom, with negatively charged electrons embedded within it.

Rutherford's Planetary Model

A model of the atom with a small, dense, positively charged nucleus at the center, with negatively charged electrons orbiting around it.