All photo electrons are not emitted with the same energy as the incident photons. Why?
Answer
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Hint: In an atom, there is an attraction between the electron and the nucleus of the atom. When the photon hits the atom some of the energy of the photon is used to overcome the attraction force and remaining energy is converted into the kinetic energy of the emitted photoelectron.
Complete step by step solution:
(i) Coulomb’s law:
The law states that there is a force in between the charges of the atom is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.
$F = {k_e}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where, $F$ is the force of attraction or repulsion between the charges in the atom, ${k_e}$ is the Coulomb’s constant, ${q_1}$ is the magnitude of the first charge,${q_2}$ is the magnitude of the second charge and $r$ is the separation between the charges.
(ii) When a photon hits the atom of the element, the photoelectron emitted from the element. But the energy of the photon is not completely converted into the kinetic energy of the photoelectron. Some of the energy is converted into the work done in removing the electron from the atom against the Coulomb’s force in the electron. This phenomenon is called the photoelectric effect. It is represented by,
Energy of photon = Maximum kinetic energy of photoelectron – The work done in removing the electron from the atom
$h\upsilon = K.{E_{\max }} - W$
Where, $h$ is Planck’s constant, $\upsilon $ is the frequency of photon wave, $K.{E_{\max }}$ is the maximum kinetic energy of the photoelectron and $W$ is the work done in removing the electron against Coulomb’s force.
Thus, the maximum kinetic energy of the emitted photoelectron, $K.{E_{\max }} = h\upsilon - W$.
Hence, the photo electrons are not emitted with the same energy as the incident photons.
Note: The kinetic energy of the photo electrons is always less than the energy of the incident photon. The characteristics of photoelectric effect are the electrons were emitted immediately, increasing the intensity of the light increased the number of photoelectrons, but not their maximum kinetic energy.
Complete step by step solution:
(i) Coulomb’s law:
The law states that there is a force in between the charges of the atom is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.
$F = {k_e}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where, $F$ is the force of attraction or repulsion between the charges in the atom, ${k_e}$ is the Coulomb’s constant, ${q_1}$ is the magnitude of the first charge,${q_2}$ is the magnitude of the second charge and $r$ is the separation between the charges.
(ii) When a photon hits the atom of the element, the photoelectron emitted from the element. But the energy of the photon is not completely converted into the kinetic energy of the photoelectron. Some of the energy is converted into the work done in removing the electron from the atom against the Coulomb’s force in the electron. This phenomenon is called the photoelectric effect. It is represented by,
Energy of photon = Maximum kinetic energy of photoelectron – The work done in removing the electron from the atom
$h\upsilon = K.{E_{\max }} - W$
Where, $h$ is Planck’s constant, $\upsilon $ is the frequency of photon wave, $K.{E_{\max }}$ is the maximum kinetic energy of the photoelectron and $W$ is the work done in removing the electron against Coulomb’s force.
Thus, the maximum kinetic energy of the emitted photoelectron, $K.{E_{\max }} = h\upsilon - W$.
Hence, the photo electrons are not emitted with the same energy as the incident photons.
Note: The kinetic energy of the photo electrons is always less than the energy of the incident photon. The characteristics of photoelectric effect are the electrons were emitted immediately, increasing the intensity of the light increased the number of photoelectrons, but not their maximum kinetic energy.
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