
An isolated metal body is illuminated with monochromatic light and is observed to become charged to a steady positive potential $1.0V$ with respect to the surrounding. The work function of the metal is $3eV$. The frequency of the incident light is.
Answer
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Hint: The question involves the concept of photoelectric effect when a monochromatic light falls on a metal surface it, after considering the work function removes electrons from the surface and the remaining energy is given as kinetic energy to the electrons. Here energy required to remove electrons will function and the energy associated with the potential of $1.0V$ developed.
This can be used to find the frequency of the incident light.
Complete step by step answer:
The energy associated with any electromagnetic wave is $E=h\upsilon $ where $h$is planck's constant whose value is $6.636\times {{10}^{-34}}{{m}^{2}}kg/s$ and $\upsilon $ is frequency of the incident light.
The photoelectric effect actually shows the particle nature of light. The above energy is associated with a photon which is the matter which actually makes up the light.
The energy associated by an electronic sphere of charge $e$ when charged to potential $V$ is $eV$.
Also the photoelectric effect equation when incident rays just remove electrons by overcoming work function and kinetic energy after removal is zero is,
$h\upsilon =work function$
But in our case it has to overcome the energy associated with the potential too, so equation becomes
$h\upsilon =work function+eV$
Work function=$3eV$
Potential=$1Volts$
$1\,electronic\,\,ch\arg e=1.6\times {{10}^{-19}}Coulomb$
Putting values
$6.636\times {{10}^{-34}}\times \upsilon =3eV+1eV$
$\Rightarrow \upsilon =\dfrac{4\times 1.6\times {{10}^{-19}}}{6.636\times {{10}^{-34}}}hertz$
$\upsilon =9.6\times {{10}^{14}}hertz$
This is the frequency of incident monochromatic light.
Note:
Work function is the minimum energy required to remove the most loosely held electron from the surface of a material. The frequency associated with the incident energy just above the work function is known as threshold frequency.
This can be used to find the frequency of the incident light.
Complete step by step answer:
The energy associated with any electromagnetic wave is $E=h\upsilon $ where $h$is planck's constant whose value is $6.636\times {{10}^{-34}}{{m}^{2}}kg/s$ and $\upsilon $ is frequency of the incident light.
The photoelectric effect actually shows the particle nature of light. The above energy is associated with a photon which is the matter which actually makes up the light.
The energy associated by an electronic sphere of charge $e$ when charged to potential $V$ is $eV$.
Also the photoelectric effect equation when incident rays just remove electrons by overcoming work function and kinetic energy after removal is zero is,
$h\upsilon =work function$
But in our case it has to overcome the energy associated with the potential too, so equation becomes
$h\upsilon =work function+eV$
Work function=$3eV$
Potential=$1Volts$
$1\,electronic\,\,ch\arg e=1.6\times {{10}^{-19}}Coulomb$
Putting values
$6.636\times {{10}^{-34}}\times \upsilon =3eV+1eV$
$\Rightarrow \upsilon =\dfrac{4\times 1.6\times {{10}^{-19}}}{6.636\times {{10}^{-34}}}hertz$
$\upsilon =9.6\times {{10}^{14}}hertz$
This is the frequency of incident monochromatic light.
Note:
Work function is the minimum energy required to remove the most loosely held electron from the surface of a material. The frequency associated with the incident energy just above the work function is known as threshold frequency.
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