Calculate the magnitude of the energy of the photon associated with light of wavelength 6057.8 Å.
Answers
Answer:
We know that, E=hv=h⋅λc
h=6.624×10−27erg−sec; c=3×1010cm/sec;
λ=6000A˚=6000×10−8cm
So, E=6×10−5(6.624×10−27)×(3×1010)=3.312×10−12erg.
Answer:
The answer to this question is 3.312×10−12erg.
Explanation:
Einstein’s Explanation of the Photoelectric Effect
Einstein resolved this problem using Planck’s revolutionary idea that light was a particle. The energy carried by each particle of light (called quanta or photon) is dependent on the light’s frequency (ν) as shown:
E = hν
Where h = Planck’s constant = 6.6261 × 10-34 Js.
Since light is bundled up into photons, Einstein theorized that when a photon falls on the surface of a metal, the entire photon’s energy is transferred to the electron.
A part of this energy is used to remove the electron from the metal atom’s grasp and the rest is given to the ejected electron as kinetic energy. Electrons emitted from underneath the metal surface lose some kinetic energy during the collision. But the surface electrons carry all the kinetic energy imparted by the photon and have the maximum kinetic energy.
We can write this mathematically as:
Energy of photon
= energy required to eject an electron (work function) + Maximum kinetic energy of the electron
E = W + KE
hv = W + KE
KE = hv – w
At the threshold frequency, ν0 electrons are just ejected and do not have any kinetic energy. Below this frequency, there is no electron emission. Thus, the energy of a photon with this frequency must be the work function of the metal.
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