Question

An electron revolving round H-nucleus in ground state absorbs \(10.2 \mathrm{eV}\) energy. Its angular momentum increases by (a) \(\frac{h}{2 \pi}\) (b) \(\frac{h}{\pi}\) (c) \(\frac{2 h}{\pi}\) (d) \(\frac{h}{4 \pi}\)

Short answer

The angular momentum increases by \( \frac{h}{2 \pi} \).

Step-by-step solution

Understanding the Problem

This exercise involves the concept of quantized angular momentum of an electron in a hydrogen atom. According to the Bohr model of the hydrogen atom, the angular momentum of an electron in an orbit is quantized and is given by the formula \( L = n \frac{h}{2 \pi} \), where n is the principal quantum number and h is the Planck's constant. When the electron moves to a higher energy state, its energy and quantum number increase.

Determining the Change in Quantum Number

The electron in the ground state (n=1) absorbs energy and transitions to a higher energy state. The energy of each state is given by \( E_n = - \frac{13.6\, \mathrm{eV}}{n^2} \). We need to find the value of n for the state to which the electron transitions after absorbing 10.2 eV. We will do this by setting the energy difference equal to the absorbed energy and solving for n.

Calculating Energy of the Initial State

We calculate the energy of the ground state (n=1) using the formula \( E_n = - \frac{13.6\, \mathrm{eV}}{n^2} \), which yields \( E_1 = -13.6\, \mathrm{eV} \).

Calculating the Energy of the Final State

Next, we add the absorbed energy to the ground state energy to find the energy of the final state: \( E_{final} = E_{initial} + \text{energy absorbed} = -13.6\, \mathrm{eV} + 10.2\, \mathrm{eV} \).

Finding the Final Energy Level

By equating the final energy to the formula for energy levels, we get \( E_n = - \frac{13.6\, \mathrm{eV}}{n^2} = E_{final} \). Solving for n gives us the final principal quantum number.

Calculating the Change in Angular Momentum

The change in angular momentum will be the difference between the angular momenta of the electron in the initial and final states, using \( \Delta L = L_f - L_i = n_f \frac{h}{2 \pi} - n_i \frac{h}{2 \pi} \). Since the initial state has n=1, we are solving for the change in angular momentum corresponding to the change in n.

Determining the Correct Option

After calculating \( \Delta L \), we compare the change in angular momentum to the given options to find the correct answer.

Key concepts

Bohr Model

The Bohr model, named after Niels Bohr, is a significant cornerstone in our understanding of atomic structure. It presents the atom as a small, positively charged nucleus surrounded by electrons that travel in circular orbits around the nucleus—much like planets orbiting the sun. However, these orbits are quantized, meaning the electrons can only orbit at certain discrete distances from the nucleus.

According to the Bohr model, each orbit corresponds to a specific energy level and is associated with a quantum number which signifies its size and energy. An interesting feature of Bohr's theory is that it specifies that as long as an electron remains in a particular orbit, it doesn't radiate energy. This contrasts with classical physics that predicts a continuously emitting electron due to acceleration. Bohr's concept of discrete orbits helps explain why we only see specific colors in the spectral lines of elements like hydrogen.

Principal Quantum Number

The principal quantum number, denoted as 'n', plays a pivotal role in the quantization of an atom's structure. It is an integer that determines the electron's energy level or shell and by extension, the likelihood of its position around an atom's nucleus.

A higher 'n' value indicates that an electron is in an energy level further away from the nucleus. Within the context of an atom's energy spectrum, higher energy levels correlate with less negative energy values. For example, an electron in the ground state has a principal quantum number of one (n=1). When energy is absorbed by an electron, 'n' increases and the electron transitions to a higher energy level, also known as 'excitation'. This change is quantized, meaning 'n' changes by whole numbers and thus, so does the corresponding angular momentum.

Planck's Constant

Planck's constant is a fundamental value in quantum mechanics denoted as 'h'. It is the proportionality constant between the minimum energy chunk—or quantum—of any particle and its frequency of emission or absorption. Put simply, it relates the energy of a photon to its electromagnetic wave frequency.

The value of Planck’s constant is approximately 6.62606957 x 10^-34 joule seconds (J s). This small number makes quantized effects prominent at the atomic level but negligible at the macroscopic scale. In the case of an atom like hydrogen, Planck’s constant quantizes the angular momentum of an electron orbiting the nucleus, with the formula for the angular momentum of an electron in a particular orbit being directly proportional to 'h'.

Energy Levels in Hydrogen Atom

In a hydrogen atom, the energy levels are the distinct states an electron can occupy, each with a unique quantum number 'n'. These levels are determined by their principal quantum numbers and can be described using the formula for the energy of each state: \( E_n = - \frac{13.6\, \mathrm{eV}}{n^2} \) where 13.6 eV represents the energy of the electron in its lowest energy state (n=1), known as the ground state.

The negative sign indicates that these are bound states with the electron's energy being lower than its potential energy when it is at infinite separation from the nucleus. When an electron transitions between these quantized energy levels, it must absorb or emit a photon with energy equal to the difference between these levels. This discrete energy change is what leads to the spectrum of lines seen in hydrogen's emission spectra and represents one of the most direct pieces of evidence of quantization in the atomic realm.