
The maximum energy of emitted photoelectrons is measured by
A) The current they produce
B) The potential they produce
C) The largest potential difference they can traverse
D) The speed with which they emerge
Answer
576.6k+ views
Hint: When a light beam of suitable frequency hits the surface of a metal, electrons are emitted from the surface of the metal. This phenomenon is known as the photoelectric effect. The emitted electrons are called photoelectrons and the current produced is called photoelectric current. The cut-off potential is the potential at which the photoelectric current will become zero.
Complete step by step solution:
For the emission of photoelectrons a minimum amount of energy is required. Thus the minimum energy required for the emission of an electron from the surface of the metal is called the work-function.
The work function $W = h{\nu _0} = \dfrac{{hc}}{{{\lambda _0}}}$
Where ${\nu _0}$ and ${\lambda _0}$ are the threshold frequency and threshold wavelength respectively.
For a particular frequency of light, if we apply a minimum retarding potential to the metal the photoelectric current will stop. This potential for which the photoelectric current is zero is called the stopping potential or the cut-off potential.
If ${V_c}$ is the cut-off potential, then the maximum kinetic energy is given by,
$K{E_{_{\max }}} = e{V_c} $
$ \Rightarrow \dfrac{1}{2}m{v^2}_{\max } = e{V_c} $
Therefore, the maximum energy of the emitted photoelectrons will be equal to the cut-off potential. The cut-off potential is the maximum potential that can be traversed by the photoelectron.
Therefore, Option (C) "The largest potential difference they can traverse" is the correct answer.
Note: Photoelectric emission is an instantaneous process. The maximum kinetic energy of the photoelectrons that are emitted will be independent of the intensity of the incident radiation. But the number of emitted photoelectrons in unit time will be proportional to the intensity of the incident radiation. The energy of the emitted photoelectrons are directly proportional to the frequency of the incident radiation.
Complete step by step solution:
For the emission of photoelectrons a minimum amount of energy is required. Thus the minimum energy required for the emission of an electron from the surface of the metal is called the work-function.
The work function $W = h{\nu _0} = \dfrac{{hc}}{{{\lambda _0}}}$
Where ${\nu _0}$ and ${\lambda _0}$ are the threshold frequency and threshold wavelength respectively.
For a particular frequency of light, if we apply a minimum retarding potential to the metal the photoelectric current will stop. This potential for which the photoelectric current is zero is called the stopping potential or the cut-off potential.
If ${V_c}$ is the cut-off potential, then the maximum kinetic energy is given by,
$K{E_{_{\max }}} = e{V_c} $
$ \Rightarrow \dfrac{1}{2}m{v^2}_{\max } = e{V_c} $
Therefore, the maximum energy of the emitted photoelectrons will be equal to the cut-off potential. The cut-off potential is the maximum potential that can be traversed by the photoelectron.
Therefore, Option (C) "The largest potential difference they can traverse" is the correct answer.
Note: Photoelectric emission is an instantaneous process. The maximum kinetic energy of the photoelectrons that are emitted will be independent of the intensity of the incident radiation. But the number of emitted photoelectrons in unit time will be proportional to the intensity of the incident radiation. The energy of the emitted photoelectrons are directly proportional to the frequency of the incident radiation.
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