/*! 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 22 A hydrogen atom initially in the... [FREE SOLUTION] | 91Ó°ÊÓ

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A hydrogen atom initially in the ground state absorbs a photon, which excites it to the \(n=4\) state. Determine the wavelength and frequency of the photon.

Short Answer

Expert verified
The wavelength of the photon is 97.1 nm, and the frequency is approximately 3.08 x 10^15 Hz.

Step by step solution

01

Calculate the Energy Levels

For a hydrogen atom, the energy levels are given by the formula: \[ E_n = -13.6 \frac{1}{n^2} \text{ eV} \]Calculate the initial energy when the atom is in the ground state \((n=1)\):\[ E_1 = -13.6 \text{ eV} \]And the energy when the atom is in the \(n=4\) state:\[ E_4 = -13.6 \frac{1}{16} = -0.85 \text{ eV} \]
02

Determine the Energy Difference

The energy of the absorbed photon corresponds to the difference between the two energy levels:\[ \Delta E = E_4 - E_1 = -0.85 \text{ eV} - (-13.6 \text{ eV}) = 12.75 \text{ eV} \]
03

Convert Energy from Electronvolts to Joules

To find the wavelength, convert the energy from electronvolts to joules. The conversion factor is \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\):\[ \Delta E = 12.75 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 2.04255 \times 10^{-18} \text{ J} \]
04

Calculate the Wavelength Using Energy

Use the energy-wavelength relationship for photons:\[ E = \frac{hc}{\lambda} \]Solve for \(\lambda\):\[ \lambda = \frac{hc}{E} \]Substitute \(h = 6.626 \times 10^{-34} \text{ Js}\) and \(c = 3 \times 10^8 \text{ m/s}\):\[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.04255 \times 10^{-18}} = 9.71 \times 10^{-8} \text{ m} = 97.1 \text{ nm} \]
05

Calculate the Frequency

Use the energy-frequency relationship:\[ E = hf \]Solve for \(f\):\[ f = \frac{E}{h} \]Substitute the known values:\[ f = \frac{2.04255 \times 10^{-18}}{6.626 \times 10^{-34}} \approx 3.08 \times 10^{15} \text{ Hz} \]

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Key Concepts

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

Photon Absorption
When we talk about photon absorption in the context of a hydrogen atom, we mean the process by which a photon of light is absorbed by the atom, causing an electron to move to a higher energy level. In the problem above, a photon is absorbed by a hydrogen atom, and it excites the electron from the ground state (where it is initially) to the fourth energy level excited state.

When a photon with the right amount of energy hits the atom, the atom can absorb the photon's energy, promoting the electron to a higher energy level. This process conserves energy, as the photon disappears, and its energy is transferred entirely to the electron.
  • This is a key mechanism behind the atom's interaction with light.
  • The absorbed photon's energy must exactly match the energy gap between the atom's initial and final energy states.
Understanding photon absorption is crucial to comprehend how light interacts with matter in different scientific contexts.
Energy Levels
Energy levels within an atom are like rungs on a ladder but defined by the allowed states an electron can occupy. For hydrogen, these levels are calculated using the formula: \[ E_n = -13.6 \frac{1}{n^2} \text{ eV} \]The negative sign indicates that the energy is lower than a free electron at rest. The ground state is the lowest energy level (n=1).

In the problem, the energy levels for n=1 and n=4 are calculated. For the ground state, \( E_1 \) is -13.6 eV, and for n=4, \( E_4 \) is -0.85 eV.
  • The transition between these levels depends on the energy provided by an incoming photon.
  • These defined energies help us understand and predict the behavior of electrons within the atom.
Recognizing how these energy levels work is vital for understanding atomic behavior.
Bohr Model
The Bohr Model of the atom was one of the first methods used to describe electron behavior in atoms, specifically hydrogen. In this model, electrons orbit the nucleus in specific paths or shells with quantized energies. This model explains why electrons occupy certain energy levels and not others.

In our example, the electron moves from a lower to a higher energy orbit upon absorbing a photon. According to Bohr's model:
  • Each orbit corresponds to a particular energy level.
  • The difference in energy between these levels explains the light absorption and emission spectrum of hydrogen.
Though more advanced models have emerged, Bohr's model remains crucial for understanding certain aspects of atomic structure and behavior.
Photon Wavelength
When an atom absorbs a photon, the wavelength of this photon can be calculated using the energy-wavelength relationship. In the solution, this is done by converting the energy difference between the atom's initial and final states into wavelength using the formula:\[ \lambda = \frac{hc}{E} \]where \( h \) is Planck's constant and \( c \) is the speed of light.

In the given problem, the energy is converted to a wavelength of 97.1 nm, belonging to the ultraviolet range.
  • Wavelength is inversely proportional to energy; higher energy photons have shorter wavelengths and vice versa.
  • By knowing the wavelength, we can determine other properties of the photon, like its frequency.
This concept is foundational for linking energy changes to visible phenomena like color and light emission.
Photon Frequency
Photon frequency refers to how often the electric field associated with a photon oscillates in a second. It is related to the energy of the photon by the equation: \[ E = hf \]where \( h \) is Planck's constant, and \( f \) is frequency. Higher energy means a higher frequency.

In the problem, the frequency of the photon that causes the transition from n=1 to n=4 is calculated to be approximately 3.08 × 10^{15} Hz.
  • Higher frequency photons carry more energy than those with lower frequencies.
  • Also, since frequency and wavelength are inversely proportional, longer wavelengths translate to lower frequencies.
Understanding photon frequency helps in numerous fields, including spectroscopy, where frequency defines the type of light interacting with the material.

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Most popular questions from this chapter

A photon with wavelength \(\lambda\) collides with a free electron. The scattered photon has a wavelength of \(\lambda^{\prime}=2 \lambda .\) If the incident photon has a wavelength of \(\lambda=1.0 \times 10^{-12} \mathrm{~m},\) through what angle is it Compton scattered?

An excited nucleus emits a gamma-ray photon with an energy of \(2.45 \mathrm{MeV}\). (a) What is the photon's energy in joules? (b) What is the photon's frequency? (c) What is the photon's wavelength? (d) How does this wavelength compare with a typical nuclear diameter of \(10^{-14} \mathrm{~m} ?\)

In a parallel universe, the value of Planck's constant is \(0.0663 \mathrm{~J} \cdot \mathrm{s}\). Assume that the physical laws and all other physical constants are the same as in our universe. In this other universe, two physics students are playing catch with a baseball. They are \(50 \mathrm{~m}\) apart, and one throws a \(0.10 \mathrm{~kg}\) ball with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\). (a) What is the uncertainty in the ball's horizontal momentum in a direction perpendicular to that in which it is being thrown if the student throwing the ball knows that it is located within a cube with volume \(1000 \mathrm{~cm}^{3}\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

In the thickness-sensitive ion microscope, a \(1.2 \mathrm{MeV}\) ion passing through a cell loses \(0.2 \mathrm{MeV}\) per \(\mu \mathrm{m}\) of thickness of the cell. If the energy of the ion can be measured to \(6 \mathrm{keV},\) what is the smallest difference in thickness that can be discerned? A. \(0.03 \mu \mathrm{m}\) B. \(0.06 \mu \mathrm{m}\) C. \(3 \mu \mathrm{m}\) D. \(6 \mu \mathrm{m}\)

If a photon of wavelength \(0.04250 \mathrm{nm}\) strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction, find (a) the change in the wavelength of this photon, (b) the wavelength of the scattered light, (c) the change in energy of the photon (is it a loss or a gain?), and (d) the energy gained by the electron.

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