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Chapter 28: Problem 23

Light of wavelength \(59 \mathrm{nm}\) ionizes a hydrogen atom that was originally in its ground state. What is the kinetic energy of the ejected electron?

Short Answer

Expert verified
The kinetic energy of the ejected electron is approximately \( 1.784 \times 10^{-17} \mathrm{J} \).

Step by step solution

01

Identify the Given Values

We are given that the light has a wavelength \( \lambda = 59 \mathrm{nm} \) (convert to meters: \(59 \times 10^{-9} \mathrm{m} \)). The electron was originally in the ground state of the hydrogen atom.
02

Calculate Energy of the Photon

The energy of a photon is calculated using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \(6.626 \times 10^{-34} \mathrm{J \cdot s}\), \( c \) is the speed of light \(3 \times 10^8 \mathrm{m/s}\) and \( \lambda \) is the wavelength. Substitute these values to find \( E \).
03

Calculate Energy of Ground State of Hydrogen

The energy of the electron in the ground state of hydrogen is \( -13.6 \mathrm{eV} \). Convert this energy into joules: \(-13.6 \mathrm{eV} \times 1.602 \times 10^{-19} \mathrm{J/eV}\).
04

Utilize Conservation of Energy

Using the conservation of energy principle, the kinetic energy of the ejected electron is given by \( KE = E_{\text{photon}} - E_{\text{initial}} \). We use the energy of the incoming photon and subtract the energy of the ground state of the hydrogen atom.
05

Calculate Kinetic Energy

Substitute the calculated values for \( E_{\text{photon}} \) and \( E_{\text{initial}} \) into the formula \( KE = E_{\text{photon}} - E_{\text{initial}} \). Convert and simplify the result of \( KE \) to obtain it in joules.

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Key Concepts

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

Hydrogen Atom
The hydrogen atom is the simplest and most basic element in the universe. It is composed of a single proton in the nucleus and one electron orbiting around it. This simplicity makes hydrogen the perfect starting point for studying atomic physics and fundamental principles of quantum mechanics. In its ground state, the electron occupies the lowest energy level possible.
In this state, the electron is closest to the nucleus, and the potential energy is minimized. This ground state energy level is crucial when calculating interactions such as those seen in the photoelectric effect, where light energy can eject an electron from an atom.
Kinetic Energy
Kinetic energy (KE) is the energy of motion. When an electron is ejected from an atom, it gains kinetic energy, allowing it to move freely. In the context of the photoelectric effect, the kinetic energy of an ejected electron can be found using the principle of energy conservation.
  • The formula for kinetic energy is given by: \( KE = E_{\text{photon}} - E_{\text{initial}} \).
  • Here, \(E_{\text{photon}}\) represents the energy of the incoming light, and \(E_{\text{initial}}\) is the initial energy of the electron (from the ground state).
The energy consequence ensures that the energy added by the photon is transformed into kinetic energy of the ejected electron.
Wavelength Calculation
To calculate the energy of a photon, it is essential to understand the role of wavelength in photon energy. The energy of a photon is inversely proportional to its wavelength.
The formula used is: \[ E = \frac{hc}{\lambda} \]
  • Here, \(h\) stands for Planck's constant \(6.626 \times 10^{-34} \text{J}\cdot\text{s}\).
  • \(c\) is the speed of light, approximately \(3 \times 10^8 \text{m/s}\).
  • \(\lambda\) is the wavelength of the light in meters.
By substituting the given wavelength into this formula, you can calculate the energy of the incoming photon involved in ionizing the atom.
Energy Conversion
Energy conversion is a fundamental concept in physics that occurs when energy changes from one form to another. In the photoelectric effect, energy from a photon is converted into kinetic energy of an electron, plus any threshold energy needed to release the electron.
This conversion can be expressed mathematically:
  • The energy from the photon \( E_{\text{photon}} \) is used to overcome the electron's binding energy \( E_{\text{initial}} \).
  • Any excess photon energy becomes the kinetic energy \( KE \) of the ejected electron.
  • This results in: \( E_{\text{photon}} = KE + E_{\text{initial}} \).
Understanding these conversions ensures a better grasp of how energy transfer works in quantum mechanics.

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Most popular questions from this chapter

Suppose that the uncertainty in position of an electron is equal to the radius of the \(n=1\) Bohr orbit, about \(0.5 \times 10^{-10} \mathrm{~m}\). Calculate the minimum uncertainty in the corresponding momentum component, and compare this with the magnitude of the momentum of the electron in the \(n=1\) Bohr orbit.

A \(2.50 \mathrm{~W}\) beam of light of wavelength \(124 \mathrm{nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is \(4.16 \mathrm{eV}\). Assume that each photon in the beam ejects an electron. (a) What is the work function (in electronvolts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

A laser used to weld detached retinas emits light with a wavelength of \(652 \mathrm{nm}\) in pulses that are \(20.0 \mathrm{~ms}\) in duration. The average power expended during each pulse is \(0.600 \mathrm{~W}\). (a) How much energy is in each pulse, in joules? In electronvolts? (b) What is the energy of one photon in joules? In electronvolts? (c) How many photons are in each pulse?

An electron in an excited state of hydrogen makes a transition from the \(n=5\) level to the \(n=2\) level. (a) Does the atom emit or absorb a photon during this process? How do you know? (b) Calculate the wavelength of the photon involved in the transition.

Light with a wavelength range of \(145-295 \mathrm{nm}\) shines on a silicon surface in a photoelectric effect apparatus, and a reversing potential of \(3.50 \mathrm{~V}\) is applied to the resulting photoelectrons. (a) What is the longest wavelength of the light that will eject electrons from the silicon surface? (b) With what maximum kinetic energy will electrons reach the anode?
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