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photoelectron question?
A sample of silver (with work function Φ=4.52 eV) is exposed to an ultraviolet light source (𝜆=200 nm),
which results in the ejection of photoelectrons. What changes will be observed if:
The silver is replaced with cobalt (Φ=5.00 eV)?
a) less energetic photoelectrons (on average)
b) fewer photoelectrons ejected
c) more energetic photoelectrons (on average)
d) no photoelectrons are emitted
e) more photoelectrons ejected
A second (identical) light source also shines on the metal?
a) less energetic photoelectrons (on average)
b) no photoelectrons are emitted
c) more photoelectrons ejected
d) more energetic photoelectrons (on average)
e) fewer photoelectrons ejected
The ultraviolet source is replaced with an X‑ray source that emits the same number of photons per unit time as the original ultraviolet source?
a) no photoelectrons are emitted
b) more energetic photoelectrons (on average)
c) more photoelectrons ejected
d) fewer photoelectrons ejected
e) less energetic photoelectrons (on average)
1 Answer
- 2 years ago
Photon energy (E) is given by:
E = hc/𝜆
where h is the planck constant; c is the speed of light and 𝜆 is the wavelength.
Here we are dealing with energy expressed in eV, so the appropriate value for (h) is
h = 4.136 × 10^-15 eV.s Plugging in values:
E = 4.136 × 10^-15 * 3.00 × 10^8 / (200 × 10^-9) = 6.204 eV
If the photon energy is less than the work function, then no photoelectrons are ejected.
If the photon energy greater than the work function, the excess energy (E - Φ) ends up as kinetic energy in the released photoelectrons.
When the silver is replaced by cobalt, the photon energy is still greater than (Φ) so you will still get the same rate of electrons ejected. However, the excess (E - Φ) is smaller, so those ejected electrons will be less energetic.
The extra light source just means more photons per unit time, so more of those less energetic electrons are ejected.
X-ray photons have a lot more energy as a result of their shorter wavelength. The new source has the same rate of productions of photons, so you will get the same numbers of ejected photoelectrons as before, but the excess (E - Φ) will be much bigger, so those ejected electrons will be fiercely more energetic.
