
In photo-electric 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 the same K.E. equal to \[h{\nu _0}\]
(B) The ejected electrons have a distribution of K.E. from zero to $h\nu - h{\nu _0}$
(C) The most energetic electrons have K.E. equal to $h\nu $
(D) All ejected electrons have K.E. equal to $h\nu $
Answer
215.1k+ views
Hint: To solve this question, we need to use Einstein's photoelectric equation. From that equation we can find out the value of the kinetic energy of the most energetic electrons.
Complete step-by-step solution:
We know that when radiation is incident on a metal surface, electrons are not ejected until the photons present in the radiation have the energy equal to or greater than the work function ${\varphi _0}$ of the metal. If the threshold frequency of the metal surface is equal to ${\nu _0}$, then the work function of the metal is given by
${\varphi _0} = h{\nu _0}$
So the photons must carry energy greater than or equal to \[h{\nu _0}\] . If the photons in the radiation have a frequency of $\nu $, then the energy of the incident radiation becomes equal to $h\nu $. Now, from the Einstein’s photoelectric equation, the kinetic energy of the ejected photoelectrons is given by
$K = h\nu - h{\nu _0}$
But we must note that this value of the kinetic energy is possessed by the electrons present at the outermost surface of the metal. This is due to the fact that the electrons embedded inside the metal have more potential energy. So they require more energy to get free. So the outermost electrons are the most energetic electrons and they have the kinetic energy equal to $h\nu - h{\nu _0}$.
As we go deeper inside the surface of the metal, the kinetic energy of the electrons decreases rapidly to zero.
Thus, the ejected electrons have a distribution of K.E. from zero to $h\nu - h{\nu _0}$.
Hence, the correct answer is option B.
Note: The photoelectric effect, which is discussed in this question, has many applications in different fields. The examples include the Photostats, which are used to generate the copies of a printed paper. Also, it is used in the photovoltaic cells, which are used in the solar panels for converting the solar energy of the sun into the electricity.
Complete step-by-step solution:
We know that when radiation is incident on a metal surface, electrons are not ejected until the photons present in the radiation have the energy equal to or greater than the work function ${\varphi _0}$ of the metal. If the threshold frequency of the metal surface is equal to ${\nu _0}$, then the work function of the metal is given by
${\varphi _0} = h{\nu _0}$
So the photons must carry energy greater than or equal to \[h{\nu _0}\] . If the photons in the radiation have a frequency of $\nu $, then the energy of the incident radiation becomes equal to $h\nu $. Now, from the Einstein’s photoelectric equation, the kinetic energy of the ejected photoelectrons is given by
$K = h\nu - h{\nu _0}$
But we must note that this value of the kinetic energy is possessed by the electrons present at the outermost surface of the metal. This is due to the fact that the electrons embedded inside the metal have more potential energy. So they require more energy to get free. So the outermost electrons are the most energetic electrons and they have the kinetic energy equal to $h\nu - h{\nu _0}$.
As we go deeper inside the surface of the metal, the kinetic energy of the electrons decreases rapidly to zero.
Thus, the ejected electrons have a distribution of K.E. from zero to $h\nu - h{\nu _0}$.
Hence, the correct answer is option B.
Note: The photoelectric effect, which is discussed in this question, has many applications in different fields. The examples include the Photostats, which are used to generate the copies of a printed paper. Also, it is used in the photovoltaic cells, which are used in the solar panels for converting the solar energy of the sun into the electricity.
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