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# Work function of the aluminium metal is 4.2eV. If two photons of, each of energy 2.5eV, are incident on its surface, will the emission of electrons take place? Justify.

Last updated date: 20th Jun 2024
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Hint: Photoelectric effect- It is the phenomenon of emission of electrons from the surface of metals, when radiations of suitable frequency fall on them. The emitted electrons are called photo-electrons and the current so produced is called photoelectric current.

Complete step by step solution:
The basic features of this effect are:
1. There is always a characteristic frequency for every metal known as threshold frequency such that the light of frequency less than ᴠₒ is incapable of emitting electrons from the emitting surface.
2. The velocity of the photo-electrons is completely independent of the intensity of incident light but depends on its frequency.
3. The rate of emission of photo-electrons is directly proportional to the intensity of the incident of light .
When the light if certain frequency(ᴠ) falls on the surface of a metal , a part of its energy is used to liberate the electrons from the surface known as the work function and rest of the energy is carried away by the emitted electrons a their kinetic energy.
This can be expressed mathematically as
$hv = \dfrac{1}{2}m{v^2} + {\phi _ \circ }$
Where ${\phi _ \circ }$ the work function of the emitting is surface and $\dfrac{1}{2}m{v^2}$ is the kinetic energy of the ejected electrons. The work function is related to threshold frequency as ${\phi _ \circ } = h{\nu _0 }$ where ${\nu _0 }$ is the minimum frequency to emit the electrons from the surface of a metal.
Now, if the frequency of the photons is then the threshold frequency, the emission of the electrons is not possible.
Thus, the ejection of electrons does not take place.

Note: The energy of the photons must be greater than or equal to work function to eject out the electrons from the surface. If the energy of photons falling on the surface is not sufficient and can not be added when the photons collide with the electrons. Thus, all the energy of the photon will be wasted and no ejection will take place.