Answer: Option (C) is the correct answer.
Explanation:
Relative between incident frequency, threshold frequency and kinetic energy is as follows.
[tex]h \nu = h \nu_{o} + \frac{1}{2}mv^{2}[/tex]
where, [tex]h \nu[/tex] = incident frequency
[tex]h \nu_{o}[/tex] = threshold frequency
K.E = [tex]\frac{1}{2}mv^{2}[/tex]
Threshold frequency is the fixed amount of energy and kinetic energy is the energy obtained due to the motion of particles of an object.
Hence, more is the frequency of photons that strikes a metal more number of electrons will come into motion. As a result, there will occur an increase in the kinetic energy of the electrons.
Thus, we can conclude that the greater the frequency of the photon that strikes a metal the greater the kinetic energy of an ejected electron.
The greater the frequency of the photon that strikes a metal, the greater the kinetic energy of an ejected electron.
The energy of incident light is directly proportional to the frequency of the incident light and it is given as follows;
E = hf
where;
The kinetic energy of the ejected electron is calculated as;
[tex]K.E = hf - \phi[/tex]
where;
Thus, from the formula given above we can conclude that, the greater the frequency of the photon that strikes a metal, the greater the kinetic energy of an ejected electron.
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