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The angular momentum (L) of an electron in a Bohr orbit is given as: (a) \(\mathrm{L}=\mathrm{nh} / 2 \pi\) (b) \(\mathrm{L}=\sqrt{[l(l+1) \mathrm{h} / 2 \pi]}\) (c) \(\mathrm{L}=\mathrm{mg} / 2 \pi\) (d) \(\mathrm{L}=\mathrm{h} / 4 \pi\)

Short Answer

Expert verified
The correct formula is (a) \( L = \frac{nh}{2\pi} \).

Step by step solution

01

Understanding the Concept

In the Bohr model of the atom, an electron orbits the nucleus and has quantized angular momentum. This quantization is due to the wave-like nature of electrons, which leads to discrete, allowable orbitals. The angular momentum is given by a specific formula in this model.
02

Applying Bohr's Angular Momentum Equation

According to the Bohr model, the angular momentum of an electron in an orbit is quantized and given by the equation: \( L = \frac{nh}{2\pi} \), where \( n \) is a principal quantum number and \( h \) is Planck's constant.
03

Identifying Correct Formula from Options

Among the given options, the formula \( L = \frac{nh}{2\pi} \) matches the one derived from Bohr's angular momentum quantization condition. Therefore, option (a) is the correct choice.

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Key Concepts

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

Angular Momentum
In the context of the Bohr model, angular momentum represents the rotational motion of an electron as it orbits the nucleus. Imagine you have a spinning toy on a string; the speed and the length of the string determine how the toy spins. Similarly, angular momentum (\( L \)) of an electron is influenced by its velocity and distance from the nucleus.
In the Bohr model, the angular momentum of an electron is quantized. This implies that an electron can only have specific, fixed angular momentum values, rather than any arbitrary amount. This was a breakthrough idea introduced by Niels Bohr, departing from classical physics, and is calculated using the formula:
  • \( L = \frac{nh}{2\pi} \)
Here, \( n \) is the principal quantum number (we'll explore this in detail later), and \( h \) is Planck's constant. The quantization of angular momentum was vital in explaining the stability of an atom, which classical physics couldn't justify.
Quantized Orbits
In the Bohr model, electrons travel in specific, defined paths around the nucleus called orbits. These paths are not random; they are quantized. But what does 'quantized' mean? It means that electrons can only exist in certain orbits that are spaced at specific distances from the nucleus. It's like stepping stones across a river: the stones exist at specific intervals, and you can only stand on them, not in between.
This concept implies that not all orbits are allowed for the electron, and only those fulfilling the quantization condition are valid. The electron's path isn't just fixed but also corresponds to specific energy levels. These quantized orbits are integral in explaining why atoms emit or absorb energy at specific wavelengths when electrons jump from one orbit to another. Such jumps between quantized orbits are responsible for the atomic spectra we observe.
Principal Quantum Number
The principal quantum number, denoted as \( n \), is a crucial factor in determining the characteristics of an electron's orbit in the Bohr model. It tells us about the size and energy of the orbits in which electrons can travel. Think of \( n \) as a floor number in an elevator, where each floor represents a different energy level. The higher you go, the more energy the electron has.
In the angular momentum equation \( L = \frac{nh}{2\pi} \), \( n \) specifies the allowed levels of angular momentum for an electron. The value of \( n \) is always a positive integer, like 1, 2, 3, etc., with \( n = 1 \) being the closest orbit to the nucleus. As \( n \) increases, the electron's orbit gets larger, and it gains more energy, affecting how it interacts with other particles and fields.
Planck's Constant
Planck's constant is a fundamental quantity that is crucial in the field of quantum mechanics, which also plays a pivotal role in the Bohr model. Represented by \( h \), it is a proportionality factor that relates the energy of a photon to its frequency. The value of Planck's constant is approximately \( 6.626 \times 10^{-34} \) Joule-seconds.
In the Bohr model, Planck's constant is integral to the quantization of angular momentum, as seen in the formula \( L = \frac{nh}{2\pi} \). This constant helps bridge the gap between the macroscopic and microscopic worlds, where energy quantization rules dominate. It provided the necessary link that allowed Bohr to explain why electrons could only exist in certain orbits. Without \( h \), the model wouldn't apply, as electron paths would not be restricted to quantized levels, contradicting observed atomic behavior.

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Most popular questions from this chapter

The electrons, identified by quantum numbers \(\mathrm{n}\) and \(l\) (i) \(\mathrm{n}=4 . l=1\) (ii) \(\mathrm{n}=4, l=0\) (iii) \(\mathrm{n}=3, l=2\) (iv) \(\mathrm{n}\) \(=3, l=1\) can be placed in order of increasing energy from the lowest to highest as: (a) (iv) \(<(\mathrm{ii})<(\mathrm{iii})<(\mathrm{i})\) (b) \((\mathrm{ii})<(\mathrm{iv})<(\mathrm{i})<(\mathrm{iii})\) (c) (i) \(<(\mathrm{iii})<(\mathrm{ii})<\) (iv) (d) (iii) \(<(\mathrm{i})<(\mathrm{iv})<(\mathrm{ii})\)

An electron is moving in Bohr's fourth orbit, its de-Broglie wavelength is \(X\). What is the circumference of the fourth orbit? (a) \(2 \lambda\) (b) \(2 / \lambda\) (c) \(3 \lambda\) (d) \(4 \lambda\)

Predict the total spin in \(\mathrm{Ni}^{2+}\) ion (a) \(\pm 5 / 2\) (b) \(\pm 3 / 2\) (c) \(\pm 1 / 2\) (d) \(\pm 1\)

An element has 2 electrons in its \(\mathrm{K}\) shell, 8 electrons in \(\mathrm{L}\) shell, 13 electrons in \(\mathrm{M}\) shell and one electron in N shell. The element is: (a) \(\mathrm{Cr}\) (b) \(\mathrm{Fe}\) (c) \(\mathrm{V}\) (d) Ti

Calculate the wavelength and energy of the radiation emitted for the electronic transition from infinity \((\infty)\) to stationary state first of the hydrogen atom. \(\left(\mathrm{R}_{\mathrm{H}}=1.09678 \times 10^{7} \mathrm{~m}^{-1}, \mathrm{~h}=6.6256 \times 10^{-34} \mathrm{Js}\right)\) (a) \(2.18 \times 10^{-21} \mathrm{~kJ}\) (b) \(3.18 \times 10^{-22} \mathrm{~kJ}\) (c) \(1.18 \times 10^{-23} \mathrm{~kJ}\) (d) \(2.18 \times 10^{-31} \mathrm{~kJ}\)

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