Problem 26 - College Physics

Question

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

Step-by-Step Solution

Answer

The kinetic energy of the ejected electron is approximately \(1.19 \times 10^{-18} \text{ J}\).

Step 1: Calculate the Energy of the Incident Photon

To find the energy of the photon, we use the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \), \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \), and \( \lambda \) is the wavelength in meters. Convert 59 nm to meters: \( 59 \text{ nm} = 59 \times 10^{-9} \text{ m} \). Hence, \[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{59 \times 10^{-9}} \approx 3.37 \times 10^{-18} \text{ J} \].

Step 2: Determine the Ionization Energy of Hydrogen

The ionization energy for a hydrogen atom in its ground state (n=1) is 13.6 eV. To use the same units as the photon energy calculated in Step 1, convert this energy into Joules: \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Therefore, \( 13.6 \text{ eV} = 13.6 \times 1.602 \times 10^{-19} \approx 2.18 \times 10^{-18} \text{ J} \).

Step 3: Calculate the Kinetic Energy of the Ejected Electron

To find the kinetic energy of the ejected electron, use the equation \( KE = E_{\text{photon}} - E_{\text{ionization}} \). From the previous steps, \( E_{\text{photon}} = 3.37 \times 10^{-18} \text{ J} \) and \( E_{\text{ionization}} = 2.18 \times 10^{-18} \text{ J} \). Thus, the kinetic energy is \( KE = 3.37 \times 10^{-18} - 2.18 \times 10^{-18} \approx 1.19 \times 10^{-18} \text{ J} \).

Key Concepts

Energy of Photon

To understand the energy of a photon, we need to recognize that photons are tiny packets of energy that make up light. The energy they carry depends on the light's wavelength. The formula to calculate this energy is given by \[ E = \frac{hc}{\lambda} \] where:

• \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \text{ J s} \).
• \( c \) is the speed of light, around \( 3.00 \times 10^8 \text{ m/s} \).
• \( \lambda \) represents the wavelength of light in meters.

The smaller the wavelength, the higher the energy of the photon. When we take a wavelength of 59 nm, which translates to \( 59 \times 10^{-9} \text{ m} \), and plug it into our equation, we can find that the energy of our specific photon is approximately \( 3.37 \times 10^{-18} \text{ J} \). This high energy is typically required to initiate processes like ionization.

Ionization Energy

Ionization energy is crucial for understanding how electrons are knocked out of atoms. It is the energy needed to remove an electron from an atom or ion. For hydrogen, especially in its ground state where the electron is closest to the nucleus, this energy is quite well-defined—13.6 eV. We often convert this to joules for calculations, using the conversion \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Hence, 13.6 eV becomes approximately \( 2.18 \times 10^{-18} \text{ J} \). This measure signifies how strongly an electron is bound to its atom and must be overcome by incoming photon energy for ionization to occur.

Kinetic Energy of Electron

The kinetic energy of an ejected electron is what remains after enough energy has been used to free the electron from the atom. This principle aligns with the photoelectric effect where the incoming photon's energy is partially used to overcome the ionization energy, with any leftover contributing to the electron's kinetic energy. We calculate this using: \[ KE = E_{\text{photon}} - E_{\text{ionization}} \] From our photon energy earlier, \( 3.37 \times 10^{-18} \text{ J} \), and our ionization energy, \( 2.18 \times 10^{-18} \text{ J} \), we find the kinetic energy is approximately \( 1.19 \times 10^{-18} \text{ J} \). This excess energy is why the electron speeds away once ionized and gives insights into processes like spectroscopy and radiation interactions.