The question asks for the maximum number of electrons that can be accommodated in the n-th shell of an atom. This is a fundamental concept in atomic structure, governed by quantum numbers and electron configuration rules.
Understanding Shells and Subshells: In an atom, electrons occupy specific energy levels called shells, denoted by the principal quantum number \(n\) (where \(n = 1, 2, 3, \ldots\)). Each shell can contain one or more subshells (s, p, d, f).
Number of Subshells in a Shell: For a given principal quantum number \(n\), there are \(n\) possible subshells. For example, for \(n=1\), there is 1 subshell (1s); for \(n=2\), there are 2 subshells (2s, 2p); for \(n=3\), there are 3 subshells (3s, 3p, 3d), and so on.
Number of Orbitals in a Subshell:
In general, for a given azimuthal quantum number \(l\), the number of orbitals is \((2l+1)\). The values of \(l\) range from \(0\) to \((n-1)\).
Maximum Electrons per Orbital: According to Pauli's Exclusion Principle, each orbital can hold a maximum of 2 electrons, provided they have opposite spins.
Calculating Total Orbitals in n-th Shell: The total number of orbitals in the n-th shell is given by \(n^2\). This can be derived by summing the number of orbitals in each subshell:
\[ \text{Total orbitals} = \sum_{l=0}^{n-1} (2l+1) \]\[ = 1 + 3 + 5 + \ldots + (2(n-1)+1) \]\[ = 1 + 3 + 5 + \ldots + (2n-1) \]\[ = n^2 \]Calculating Maximum Electrons in n-th Shell: Since each orbital can hold a maximum of 2 electrons, the maximum number of electrons in the n-th shell is twice the total number of orbitals in that shell.
\[ \text{Maximum electrons} = 2 \times (\text{Total orbitals}) \]\[ = 2 \times n^2 \]\[ = 2n^2 \]C) 2n² is the correct formula for the maximum number of electrons in the n-th shell, based on the principles of quantum mechanics and electron configuration.
The question asks to identify the experiment that led to the discovery of the atomic nucleus. We need to recall the key findings of each experiment listed in the options.
Based on these points, Rutherford's alpha scattering experiment is the one that demonstrated the existence of the atomic nucleus.
Correct Option: D) Rutherford's alpha scattering experiment
Rutherford's gold foil experiment, also known as the Geiger-Marsden experiment, was crucial in understanding the structure of the atom. By observing how alpha particles interacted with a thin gold foil, Rutherford deduced key properties about the atomic nucleus.
Therefore, the most common observation was that alpha particles passed straight through the foil.
Correct Option: A) Passed straight through the foil
The question asks for the formula for the energy of an electron in the n-th orbit of a hydrogen atom. This is a fundamental concept in atomic physics, specifically derived from Bohr's model of the hydrogen atom.
Correct Option: B) -13.6/n² eV
The radius of the Bohr orbit for a hydrogen-like atom can be calculated using the Bohr model's formula. For the first Bohr orbit of a hydrogen atom, we need to substitute the principal quantum number \(n=1\) and the atomic number \(Z=1\) into the general formula for the Bohr radius.
The general formula for the radius of the \(n\)-th Bohr orbit for a hydrogen-like atom is given by:
\[ r_n = \frac{n^2 a_0}{Z} \]where:
For the first Bohr orbit of a hydrogen atom:
Substitute these values into the formula:
\[ r_1 = \frac{(1)^2 a_0}{1} = a_0 \]Therefore, the radius of the first Bohr orbit of a hydrogen atom is equal to the Bohr radius \(a_0\).
The standard value for the Bohr radius is \(0.529 \text{ Å}\).
D) 0.529 Å — The radius of the first Bohr orbit of a hydrogen atom is defined as the Bohr radius, \(a_0\), which has a value of approximately \(0.529 \text{ Å}\).
The question asks for the ground state energy of a hydrogen atom. This is a fundamental concept in atomic physics, specifically related to Bohr's model of the atom or quantum mechanics. The energy levels of a hydrogen-like atom can be calculated using a specific formula.
Recall the formula for energy levels in a hydrogen atom: The energy \(E_n\) of an electron in the \(n\)-th orbit of a hydrogen atom is given by the formula:
\[ E_n = - \frac{13.6}{n^2} \text{ eV} \]where \(n\) is the principal quantum number.
Identify the ground state: The ground state corresponds to the lowest energy level, which means the principal quantum number \(n=1\).
Substitute \(n=1\) into the formula:
\[ E_1 = - \frac{13.6}{(1)^2} \text{ eV} \]\[ E_1 = - \frac{13.6}{1} \text{ eV} \]\[ E_1 = -13.6 \text{ eV} \]Compare with the given options: The calculated ground state energy is \(-13.6 \text{ eV}\).
C) -13.6 eV — This is the correct value for the ground state energy of a hydrogen atom, derived from the Bohr model and confirmed by quantum mechanics.