Photoelectron spectroscopy (PES) is a technique based on the photoelectric effect.In PES, monochromatic light is directed onto a sample, causing electrons to beemitted. The difference between the energy of the photons and the kinetic energy ofthe electrons corresponds to the ionization energy. Suppose that a PES experiment isperformed, in which certain gas X is irradiated with ultraviolet light of photon energy12.1 eV. The kinetic energy of the emitted electrons is measured to be 4.8 eV. Whatis the first ionization energy of X?O-16.9 eV12.1 eV16.9 eVO-7.3 eV7.3 eV

The Photon energy or Potential energy= 12.1eV The Kinetic energy= 4.8eV

Answered: Photoelectron spectroscopy (PES) is a…

Similar questions

Photoelectrons with a maximum kinetic energy of 7.95 eV are emitted from a metal when it is illuminated by ultraviolet radiation with a wavelength of 1.25 × 10-7 m. (i) Calculate the energy of the incident photons in electronvolts. (ii) Calculate the wavelength of the radiation (in nanometres) corresponding to the lowest energy photons that can free electrons from the metal. (iii) How do you explain the fact that, when infrared radiation is shone on this metal, no photoelectrons are emitted?
Suppose a photon has an energy of 4 electron volts (eV). What is the energy (in eV) of a second photon with a frequency that is 70% that of the first one ? {This means the frequency is less . For example, 10 % the original frequency equals 10/100 times the initial frequency, or (10/100) times the initial frequency ]
When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV. (a) What is the work function of the metal surface? (b) What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0 nm falls on the same surface?
The mathematical equation for studying the photoelectric effect is hν = W + 1/2 meu2 where ν is the frequency of light shining on the metal; W is the energy needed to remove an electron from the metal; and me and u are the mass and speed of the ejected electron, respectively. In an experiment, a student found that a maximum wavelength of 351 nm is needed to just dislodge electrons from a metal surface. Calculate the velocity (in m/s) of an ejected electron when the student employed light with a wavelength of 283 nm.
Suppose a star with radius 8.57 × 108 m has a peak wavelength of 680 nm in the spectrum of its emitted radiation. (a) Find the energy of a photon with this wavelength. J/photon (b) What is the surface temperature of the star? K (c) At what rate is energy emitted from the star in the form of radiation? Assume the star is a blackbody (e = 1). W (d) Using the answer to part (a), estimate the rate at which photons leave the surface of the star. photons/s
Ultraviolet light with a single wavelength and with an intensity of 550 W/m² is incident normally on the surface of a metal that has a work function of 3.44 eV. Photoelectrons are emitted with a maximum speed of 420 km/s. Find the maximum possible rate of photoelectron emission from 1 cm² of the surface by imagining that every photon produces one photoelectron.
For light with a wavelength of 350 nm and with an intensity of /= 10-8 W/m², what is the number of photons/(m²s) in the light beam?
The work function of caesium metal is 2.14 eV. When light of frequency 6 x 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum kinetic energy of the emitted electrons,
A photon having a wavelength of 182 nm strikes the surface of a metal sheet having a threshold frequency (νo) of 9.27 x 1014 s-1. Calculate the velocity of the ejected electron in SI units AND in units of miles per hour. Hint: Combine equations 1.3 (pg 16 of textbook) and 3.6 (pg. 94). Report the units of your answer.