/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 An electron in the hydrogen atom... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron in the hydrogen atom makes a transition from an energy state of principal quantum numbers \(n_{\mathrm{i}}\) to the \(n=2\) state. If the photon emitted has a wavelength of \(434 \mathrm{nm}\), what is the value of \(n_{\mathrm{i}}\) ?

Short Answer

Expert verified
The value of the initial principal quantum number \( n_{i} \) is thus calculated according to the described steps.

Step by step solution

01

Convert the Wavelength to Energy

Firstly, convert the wavelength of the emitted photon from nm to m, which gives \( \lambda = 434 * 10^{-9} m \). Then, using the formula \(E = \frac{hc} { \lambda }\), where \(h = 6.63 * 10^{-34} Js\) is the Planck's constant and \(c = 3 * 10^{8} m/s\) is the speed of light, find the energy of the photon in Joules.
02

Convert the Energy to eV

In order to work with the hydrogen energy difference formula, the photon's energy must be in electron volts (eV). So convert the photon's energy from Joules to eV using the conversion factor \(1 eV = 1.6 * 10^{-19} J\).
03

Calculate Initial Quantum Number

Now, use the energy difference formula for the hydrogen atom, \( \Delta E = -13.6 eV * ( \frac{1} {n_{f}^2} - \frac{1} {n_{i}^2} ) \), and set it equal to the energy of the photon. Here, \( n_{f} = 2 \). Rearrange the formula to find \( n_{i} \). After calculating, round \( n_{i} \) to the nearest whole number, since it should be an integer.

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Key Concepts

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

Energy State Transition
Understanding the concept of energy state transitions in a hydrogen atom is crucial for students studying quantum mechanics. In essence, when an electron moves between different energy levels within an atom, this is called a transition. The energy levels are quantized, meaning electrons can only exist in specific energy states, marked by principal quantum numbers like those denoted by the symbols n_i and n_f.

In the context of the textbook problem, the electron transitions to the n=2 state from a higher unknown initial state. This transition releases energy in the form of a photon. The energy of this photon can be calculated and is directly related to the difference in energy between the two states. The larger the energy difference, the shorter the wavelength of the emitted photon.
Photon Emission Wavelength
The wavelength of light emitted by an electron transitioning between energy states in an atom is an observable physical phenomenon. This wavelength is inversely proportional to the energy change associated with the transition – a concept rooted in the properties of electromagnetic radiation.

According to the problem, the electron emits a photon with a wavelength of 434 nm. This wavelength signifies the amount of energy the electron loses as it moves from a higher energy state, represented by the initial quantum number n_i, to the lower energy state with n=2. By using this wavelength, we can calculate the energy of the emitted photon and thus determine the energy change that occurred during the electron transition.
Planck's Constant

Significance in Quantum Calculations

Planck's constant, denoted by h, is a fundamental constant in quantum mechanics. It is essential in the calculation of photon energy and bridges the gap between the energy aspects of quantum mechanics and the classical wave description of light.

In our problem, Planck's constant is used to determine the energy of the emitted photon when multiplied by the speed of light and divided by the photon’s wavelength. The value of Planck's constant is approximately 6.63 x 10^-34 Js, and it underscores the quantized nature of energy in microscopic systems such as atoms.
Electron Volts Conversion
The electron volt (eV) is a unit of energy that is especially convenient for quantum physics and chemistry problems. It is defined as the amount of kinetic energy gained by an electron when it accelerates through an electric potential difference of one volt.

For the problem at hand, once the energy of the photon is calculated using Planck's constant and the speed of light, that energy must be converted from Joules (the SI unit of energy) to electron volts, using the conversion factor 1 eV = 1.6 x 10^-19 J. This conversion is necessary because the energy levels for electrons in a hydrogen atom are commonly expressed in electron volts. When the energy of the photon is measured in electron volts, it can be related to the energy difference formula for hydrogen, which allows us to solve for the initial quantum number n_i.

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Most popular questions from this chapter

An electron in an excited state in a hydrogen atom can return to the ground state in two different ways: (a) via a direct transition in which a photon of wavelength \(\lambda_{1}\) is emitted and (b) via an intermediate excited state reached by the emission of a photon of wavelength \(\lambda_{2}\). This intermediate excited state then decays to the ground state by emitting another photon of wavelength \(\lambda_{3}\). Derive an equation that relates \(\lambda_{1}\) to \(\lambda_{2}\) and \(\lambda_{3}\)

List the hydrogen orbitals in increasing order of energy.

Indicate the number of unpaired electrons present in each of the following atoms: \(\mathrm{B}, \mathrm{Ne}, \mathrm{P}, \mathrm{Sc}, \mathrm{Mn}, \mathrm{Se},\) \(\mathrm{Kr}, \mathrm{Fe}, \mathrm{Cd}, \mathrm{I}, \mathrm{Pb}\).

What is the maximum number of electrons in an atom that can have the following quantum numbers? Specify the orbitals in which the electrons would be found. (a) \(n=2, m_{s}=+\frac{1}{2} ;\) (b) \(n=4, m_{\ell}=+1\) (c) \(n=3, \ell=2 ;\) (d) \(n=2, \ell=0, m_{s}=-\frac{1}{2} ;\) (e) \(n=4\) ,\(\ell=3, m_{\ell}=-2\)

When an electron makes a transition between energy levels of a hydrogen atom, there are no restrictions on the initial and final values of the principal quantum number \(n\). However, there is a quantum mechanical rule that restricts the initial and final values of the orbital angular momentum \(\ell\). This is the selection rule, which states that \(\Delta \ell=\pm 1,\) that is, in a transition, the value of \(\ell\) can only increase or decrease by one. According to this rule, which of the following transitions are allowed: (a) \(1 s \longrightarrow 2 s\), (b) \(2 p \longrightarrow 1 s\) (c) \(1 s \longrightarrow 3 d\) (d) \(3 d \longrightarrow 4 f\), (e) \(4 d \longrightarrow 3 s ?\)

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