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If \(n\) and \(l\) are, respectively, the principal and azimuthal quantum numbers, then the expression for calculating the total number of electrons in any energy level is (a) \(\sum_{l=1}^{l=n} 2(2 l+1)\) (b) \(\sum_{l=1}^{l=n-1} 2(2 l+1)\) (c) \(\sum_{l=0}^{l=n+1} 2(2 l+1)\) (d) \(\sum_{l=0}^{l=n-1} 2(2 l+1)\)

Short Answer

Expert verified
The total number of electrons in any energy level is given by option (d) \(\sum_{l=0}^{l=n-1} 2(2l+1)\).

Step by step solution

01

Understanding the quantum numbers

The principal quantum number, denoted as 'n', specifies the energy level of an electron in an atom and can take positive integer values (n=1,2,3,...). The azimuthal quantum number, denoted as 'l', defines the subshell (s, p, d, f,...) and can take on integer values from 0 to n-1 for each value of n. Each subshell can hold a maximum of \(2(2l+1)\) electrons.
02

Identifying the correct option

Since the azimuthal quantum number 'l' ranges from 0 to n-1 for a given 'n', the expression for the total number of electrons in any energy level must include a summation from \(l=0\) to \(l=n-1\). Each subshell contributes \(2(2l+1)\) electrons. Thus, the correct expression for the total number of electrons in an energy level will have the lower limit of the summation as 0 and the upper limit as n-1.
03

Selecting the correct expression

Considering the structure of the options and the information from the previous steps, the correct expression for the total number of electrons in any energy level must sum the number of electrons in each subshell from \(l=0\) to \(l=n-1\), which is given by option (d): \(\sum_{l=0}^{l=n-1} 2(2l+1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Principal Quantum Number
The principal quantum number, represented by the symbol 'n', is an integral part of quantum mechanics that defines the energy level of an electron within an atom. Think of it as the 'floor' number in a hotel where the energy level increases with the floor number. The value of 'n' starts at 1, indicating the closest energy level to the nucleus, and increases as you move further away, corresponding to higher energy levels.

Each energy level can hold more electrons as 'n' gets larger, according to a specific formula based on quantum theory. Knowing the principal quantum number is crucial for understanding how atoms are structured and how they bond with other atoms, as well as for predicting the chemical properties of an element.
Azimuthal Quantum Number
The azimuthal quantum number, denoted by 'l', relates to the shape of the electron cloud within an energy level, which corresponds to different subshells labeled as s, p, d, and f. Each 'l' value ranges from 0 up to (n-1), where 'n' is the principal quantum number. For instance, if 'n' is 3, 'l' can be either 0 (s-subshell), 1 (p-subshell), or 2 (d-subshell).

This quantum number is important for understanding the complex behaviors of electrons, such as chemical bonding and magnetic properties. The subshells are like different types of rooms on the same 'floor' (energy level), and each type (s, p, d, f) can accommodate a different number of electrons.
Electron Configuration
Electron configuration is essentially the address book for electrons within an atom. It is a detailed map that tells us where an electron can be found within its energy levels (floors) and subshells (rooms). The specific sequence in which electrons are arranged in those subshells determines the atom's chemical behavior.

Writing out an electron configuration involves using the quantum numbers to jot down a structured 'address label' for electrons, and following a set of rules, such as the Pauli exclusion principle and Hund's rule, to ensure each electron has a unique place within the atom. This configuration helps chemists and physicists predict how atoms will interact in chemical reactions and what types of bonds they will form.
Quantum Chemistry
Quantum chemistry is a field that combines the principles of quantum physics with chemical systems. It's about understanding how quantum mechanics governs the behavior of atoms and molecules. At its core is the use of quantum numbers, like the principal and azimuthal numbers, and understanding electron configurations to explore and explain the nature of chemical bonds, reaction dynamics, and molecular properties.

In quantum chemistry, scientists use the 'language' of mathematics and computational models to delve deep into the microscopic world, revealing the energetic and structural secrets of matter. By deciphering these quantum 'codes', researchers can predict the outcomes of chemical reactions, design new materials, and even develop novel drugs to tackle diseases, making it a truly interdisciplinary and impactful branch of science.

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Most popular questions from this chapter

For an electron in a hydrogen atom, the wave function is given by \(\psi_{1 \mathrm{~s}}\) \(=(\pi / \sqrt{2}) e^{-r / a_{0}}\), where \(a_{0}\) is the radius of first Bohr's orbit and \(r\) is the distance from the nucleus with which probability of finding electron varies. What will be the ratio of probabilities of finding electrons at the nucleus to first Bohr's orbit \(a_{\mathrm{o}} ?\) (a) 0 (b) \(e\) (c) \(e^{2}\) (d) \(\frac{1}{e^{2}}\)

A compound of vanadium has a magnetic moment of \(1.73 \mathrm{BM}\). The electronic configuration of vanadium ion in the compound is (a) \([\mathrm{Ar}] 3 \mathrm{~d}^{2}\) (b) \([\mathrm{Ar}] 3 \mathrm{~d}^{1}\) (c) \([\mathrm{Ar}] 3 \mathrm{~d}^{3}\) (d) [Ar]4s

Uncertainty in the position of an electron (mass \(=9.1 \times 10^{-31} \mathrm{~kg}\) ) moving with a velocity \(300 \mathrm{~m} / \mathrm{s}\), accurate up to \(0.001 \%\), will be \(\left(h=6.3 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\) (a) \(5.76 \times 10^{-2} \mathrm{~m}\) (b) \(1.92 \times 10^{-2} \mathrm{~m}\) (c) \(3.84 \times 10^{-2} \mathrm{~m}\) (d) \(19.2 \times 10^{-2} \mathrm{~m}\)

Which of the following orbital does not exist according to quantum theory? (a) \(5 \mathrm{~g}\) (b) \(4 \mathrm{f}\) (c) \(5 \mathrm{~h}\) (d) \(6 \mathrm{~h}\)

The ionization energy of a hydrogen-like atom is \(14.4 \mathrm{eV} .\) The amount of energy released when electron jumps from the fourth orbit to the first orbit in this atom, is (a) \(13.5 \mathrm{eV}\) (b) \(10.8 \mathrm{eV}\) (c) \(0.9 \mathrm{eV}\) (d) \(12.75 \mathrm{eV}\)

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