Problem 70
Question
The energy of second Bohr orbit of the hydrogen atom is \(-328 \mathrm{~kJ} \mathrm{~mol}^{-1}\), hence the energy of fourth bohr orbit would be (a) \(-164 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-41 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-82 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-Step Solution
Verified Answer
The energy of the fourth Bohr orbit is \(-82 \text{ kJ/mol}\).
1Step 1: Understand the Bohr's energy equation
Bohr's model of the hydrogen atom expresses the energy of an electron in a given orbit as: \(E_n = \frac{-1312}{n^2} \text{ kJ/mol}\), where \(n\) is the principal quantum number (orbit number).
2Step 2: Calculate energy of second orbit
According to the given information, the energy of the second orbit (\(n=2\)) is \(-328 \text{ kJ/mol}\). Check consistency with Bohr's formula: \(E_2 = \frac{-1312}{2^2} = \frac{-1312}{4} = -328 \text{ kJ/mol}\). The given energy for the second orbit is correct.
3Step 3: Calculate energy of the fourth orbit
Use Bohr's formula to find the energy of the fourth orbit (\(n=4\)): \(E_4 = \frac{-1312}{4^2} = \frac{-1312}{16}\). Calculate this to find the energy in the fourth orbit.
4Step 4: Compute the Result
Perform the division: \(E_4 = \frac{-1312}{16} = -82 \text{ kJ/mol}\). Thus, the energy of the fourth Bohr orbit is \(-82 \text{ kJ/mol}\).
Key Concepts
Principal Quantum NumberEnergy LevelsHydrogen Atom
Principal Quantum Number
The principal quantum number, denoted by \(n\), plays a crucial role in defining the structure and energy levels of an atom. It is an integer that determines the size and energy of an atom's orbitals. In the Bohr model of the hydrogen atom, the principal quantum number represents the electron's orbit, or shell, around the nucleus. These orbits are labeled as \(n=1, 2, 3,\) and so on.
- The value of \(n\) increases with the distance from the nucleus, meaning higher \(n\) values represent orbits further out.
- The smallest \(n\) value, \(n=1\), is called the ground state, while larger values are referred to as excited states.
- Each orbit corresponds to a specific energy level, with energy increasing as \(n\) increases.
Energy Levels
Energy levels within an atom correlate directly with the principal quantum number \(n\). Each energy level is quantized, meaning it has a specific value and electrons can only exist in one of these distinct levels. In the Bohr model, energy levels are often visualized as rungs on a ladder. Electrons moving between these levels emit or absorb energy as light, a concept known as the quantum leap.
- The lowest energy level (ground state) is the most stable and occurs at \(n=1\).
- As \(n\) increases, energy levels become closer to one another.
- Energy required to remove an electron from an atom, ionization energy, is an important aspect of these levels.
Hydrogen Atom
The hydrogen atom, the simplest atom consisting of just one proton and a single electron, is foundational in the study of atomic structure and quantum mechanics. Niels Bohr's model gave insight into how energy levels work within such atoms. His model proposed that electrons move in specific circular orbits, or shells, around the nucleus.
- The model successfully explains the spectral lines of hydrogen seen in its emission spectrum.
- The hydrogen atom's energy levels can be calculated using Bohr's energy equation: \(E_n = \frac{-1312}{n^2} \text{ kJ/mol}\).
- Despite its limitations, Bohr's model was pivotal in advancing atomic theory and laid the groundwork for modern quantum mechanics.
Other exercises in this chapter
Problem 68
Given: the mass of electron is \(9.11 \times 10^{-31} \mathrm{~kg}\) Planck constant is \(6.626 \times 10^{-34} \mathrm{Js}\), the uncertainty involved in the m
View solution Problem 69
The energy ratio of a photon of wavelength \(3000 \AA\) and \(6000 \AA\) is (a) \(1: 1\) (b) \(2: 1\) (c) \(1: 2\) (d) \(1: 4\)
View solution Problem 71
The de Broglie wavelength associated with a particle of mass \(10^{-6} \mathrm{~kg}\) moving with a velocity of \(10 \mathrm{~ms}^{-1}\) is (a) \(6.63 \times 10
View solution Problem 72
In hydrogen atom, energy of first excited state is \(-3.4\) \(\mathrm{eV}\). The kinetic energy of the same orbit of hydrogen atom would be (a) \(+3.4 \mathrm{e
View solution