
In the photoelectric effect, when photons of energy$h\nu $are incident on a metal surface, electrons are emitted with some kinetic energy. It is possible to say that
A) All ejected electrons have same K.E equal to $h{\nu _o}$
B) The ejected electrons have a distribution of K.E from 0 to $h\nu - h{\nu _o}$
C) The most energetic electrons have K.E equal to $h\nu $
D) All ejected electrons have K.E equal to $h{\nu _o}$
Answer
573.6k+ views
Hint-When the light is incident on a metal surface, electrons are emitted. This phenomenon is called photoelectric effect as the electrons are emitting on the incident photons. The photon of energy $h\nu $ is incident on the metal surface. This energy is used to remove the electron from its orbit of a particular energy level, also providing the kinetic energy for it.
Complete step by step answer:
(i) Consider the light source having the photons of energy $h\nu $ is illuminating on the metal surface. The photon of energy $h\nu $ is removing the electrons from its orbit of energy $h{\nu _o}$. This ejected or removed electron now having the kinetic energy K.E.
(ii) Hence the photon’s energy $h\nu $ is used to eject the electron of particular energy and also provide kinetic energy to it for its motion.
$\therefore h\nu = h{\nu _o} + K.E$
(iii) Now we have to find the kinetic energy of the electrons, therefore
$K.E = h\nu - h{\nu _o}$
(v) Sometimes the complete energy of the photon is used for ejecting the electrons. In this case, the kinetic energy is zero. Therefore we can say that the kinetic energy of the ejected electrons can be ranged from zero to $h\nu - h{\nu _o}$
(iv)From the above explanation, we can assure that the correct option is B.
Note:The photoelectric effect was first discovered by Albert Einstein and he also won the Nobel Prize for it. The electrons ejected after the incident of photons are called photoelectrons. And the kinetic energy of these photo electrons increases with the light frequency. After finding the light is a particle, the kinetic energy of the photo electrons remains constant when light amplitude increases.
Complete step by step answer:
(i) Consider the light source having the photons of energy $h\nu $ is illuminating on the metal surface. The photon of energy $h\nu $ is removing the electrons from its orbit of energy $h{\nu _o}$. This ejected or removed electron now having the kinetic energy K.E.
(ii) Hence the photon’s energy $h\nu $ is used to eject the electron of particular energy and also provide kinetic energy to it for its motion.
$\therefore h\nu = h{\nu _o} + K.E$
(iii) Now we have to find the kinetic energy of the electrons, therefore
$K.E = h\nu - h{\nu _o}$
(v) Sometimes the complete energy of the photon is used for ejecting the electrons. In this case, the kinetic energy is zero. Therefore we can say that the kinetic energy of the ejected electrons can be ranged from zero to $h\nu - h{\nu _o}$
(iv)From the above explanation, we can assure that the correct option is B.
Note:The photoelectric effect was first discovered by Albert Einstein and he also won the Nobel Prize for it. The electrons ejected after the incident of photons are called photoelectrons. And the kinetic energy of these photo electrons increases with the light frequency. After finding the light is a particle, the kinetic energy of the photo electrons remains constant when light amplitude increases.
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