The work function for a metal is 4.31 eV. Ultraviolet light of wavelength 125 nm is incident on the clean surface of this metal. What is the photon energy of the ultraviolet light in units of eV?Answer for part 1What is the maximum kinetic energy (in eV) of the ejected photoelectrons?Answer for part 2What stopping voltage would be required to arrest the current of photoelectrons?
Question
The work function for a metal is 4.31 eV. Ultraviolet light of wavelength 125 nm is incident on the clean surface of this metal. What is the photon energy of the ultraviolet light in units of eV?Answer for part 1What is the maximum kinetic energy (in eV) of the ejected photoelectrons?Answer for part 2What stopping voltage would be required to arrest the current of photoelectrons?
Solution
Part 1:
The energy of a photon can be calculated using the formula E = hc/λ, where h is Planck's constant (4.1357 x 10^-15 eV.s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the light.
First, we need to convert the wavelength from nm to m: 125 nm = 125 x 10^-9 m.
Then, we can substitute these values into the formula:
E = (4.1357 x 10^-15 eV.s x 3 x 10^8 m/s) / (125 x 10^-9 m)
E = 9.93 eV
So, the photon energy of the ultraviolet light is 9.93 eV.
Part 2:
The maximum kinetic energy of the ejected photoelectrons can be calculated using the formula K.E. = E - W, where E is the energy of the incident light and W is the work function of the metal.
K.E. = 9.93 eV - 4.31 eV = 5.62 eV
So, the maximum kinetic energy of the ejected photoelectrons is 5.62 eV.
Part 3:
The stopping voltage can be calculated using the formula V = K.E./e, where K.E. is the maximum kinetic energy of the photoelectrons and e is the charge of an electron (1.6 x 10^-19 C).
V = 5.62 eV / 1.6 x 10^-19 C = 3.51 x 10^-19 J / 1.6 x 10^-19 C = 2.19 V
So, a stopping voltage of 2.19 V would be required to arrest the current of photoelectrons.
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