Chapter 4: Problem 12
The critical potential of hydrogen is \(13.65 \mathrm{eV}\). Calculate the wavelength of the radiation emitted by hydrogen atom bombarded by an electron of corresponding energy.
Short Answer
Expert verified
The wavelength of the emitted radiation is 91 nm.
Step by step solution
01
- Understand the problem
The goal is to find the wavelength of the radiation emitted by a hydrogen atom when it is bombarded by an electron with an energy of 13.65 eV. The critical potential energy provided allows us to use the relationship between energy and wavelength.
02
- Use Energy-Wavelength Relationship
The energy of a photon is given by the formula: \[ E = \frac{hc}{u} \]where:- \( E \) is the energy in electron volts (13.65 eV)- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s} \))- \( c \) is the speed of light (\( 3.00 \times 10^8 \text{ m/s} \))- \( u \) is the frequency of the radiationWe know that \( E = h u \).
03
- Convert Energy to Joules
First, convert the energy from electron volts to joules: \[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]This means: \[ E = 13.65 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 2.18643 \times 10^{-18} \text{ J} \]
04
- Calculate the Wavelength
Using the formula \( E = \frac{hc}{u} \) and substituting \( E \) and solving for \( u \), we get: \[ u = \frac{c}{\text{wavelength}} \rightarrow \text{wavelength} = \frac{hc}{E} \]Substitute the known values: \[ \text{wavelength} = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m/s}}{2.18643 \times 10^{-18} \text{ J}} \]\[ \text{wavelength} = 9.10 \times 10^{-8} \text{ m} \]\( \text{wavelength} = 91 \text{ nm} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Energy-Wavelength Relationship
To find the wavelength of radiation emitted by a hydrogen atom, we start by understanding how energy and wavelength are related. This relationship is described by the formula: \[ E = \frac{hc}{\text{wavelength}} \] Here:
- E is the energy of the photon (in joules).
- c is the speed of light (around \(3.00 \times 10^8 \text{ m/s})\).
- h is Planck's constant \(6.626 \times 10^{-34} \text{ J s})\).
Electron Volts to Joules Conversion
In physics, energy can be measured in different units. Electron volts (eV) is a common unit when discussing atomic and subatomic particles. However, for calculations involving Planck's constant and the speed of light, we use joules (J). We convert electron volts to joules using the relationship:\[1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]To convert a given energy from eV to J, multiply by this factor. For example, for the energy of 13.65 eV given in the exercise, we have:\[ 13.65 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 2.18643 \times 10^{-18} \text{ J} \] Converting energy into joules ensures that our calculations remain consistent and aligns with the constants we use.
Planck's Constant
Planck's constant, denoted as h, is a fundamental constant in quantum mechanics. It relates the energy of a photon to the frequency of its associated electromagnetic wave via the equation:\[ E = h u \] Here:
- h is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s}\)).
- E is the energy (in joules).
- ν (nu) is the frequency of the radiation (in Hz).
Critical Potential Energy
Critical potential energy is the minimum energy required to ionize an electron from an atom. For hydrogen, this value is 13.65 eV. When an electron with this energy collides with a hydrogen atom, it excites the atom, causing it to emit radiation. This energy input leads to the emission of a photon with a specific wavelength. This principle is foundational in understanding emission spectra in quantum mechanics.
- Knowing the critical potential helps us calculate the exact wavelength of the resulting radiation.
- The energy is directly used in the energy-wavelength relationship to find the wavelength of emitted photons.