
A silver sphere (work function 4.6eV) is suspended in a vacuum chamber by an insulating thread Ultraviolet light of wavelength 0.2μm strike. The maximum electric potential of the sphere will be
A. 4.6V
B. 6.2V
C. 1.6V
D. 3.2V
Answer
164.4k+ views
Hint: When an electromagnetic radiation of a fixed wavelength and frequency falls on a metal surface, electrons are emitted having a fixed value of kinetic energy corresponding to that wavelength. We will apply the equation of photoelectric effect to find maximum electric potential when light of given wavelength falls on the surface of a silver sphere.
Formula used:
We will be using the formula
$U = eV$ and $U = \dfrac{{hc}}{\lambda } - \Phi $
Where $\lambda = $Wavelength and$\Phi = $Work function of metal
Complete answer:
Work function is the least amount of energy necessary to remove an electron from the surface of a certain material, usually a metal, to infinity.
The photoelectric effect is described as the emission of electrons when electromagnetic radiation hits a metal surface. Electrons emitted during this process are called photoelectrons. We can say that in photoelectric effect, electrically charged particles, are released from or within a material when it absorbs electromagnetic radiation. The ejection of electrons from a metal surface plate when light strikes it is how this process is explained.
Substituting the values from question we get,
$U = \dfrac{{12400eV\mathop A\limits^ \circ }}{{2000\mathop A\limits^ \circ }} - 4.6eV$
$ \Rightarrow U = 6.2eV - 4.6eV = 1.6eV$
Now $U = 1.6eV = e{V^{\prime}}$
So ${V^{\prime}} = 1.6V$
The maximum electric potential of the sphere came out to be 1.6eV. Option C is correct.
Note:Photoelectrons cannot be produced by light waves of any frequency. The photoelectric effect is brought on by light that is below a specific cut-off frequency. The more intense the light, above the cut-off voltage, the higher the kinetic energy the released photons would have. The quantity of photons released is inversely related to the frequency of incident light for a constant magnitude of light intensity.
Formula used:
We will be using the formula
$U = eV$ and $U = \dfrac{{hc}}{\lambda } - \Phi $
Where $\lambda = $Wavelength and$\Phi = $Work function of metal
Complete answer:
Work function is the least amount of energy necessary to remove an electron from the surface of a certain material, usually a metal, to infinity.
The photoelectric effect is described as the emission of electrons when electromagnetic radiation hits a metal surface. Electrons emitted during this process are called photoelectrons. We can say that in photoelectric effect, electrically charged particles, are released from or within a material when it absorbs electromagnetic radiation. The ejection of electrons from a metal surface plate when light strikes it is how this process is explained.
Substituting the values from question we get,
$U = \dfrac{{12400eV\mathop A\limits^ \circ }}{{2000\mathop A\limits^ \circ }} - 4.6eV$
$ \Rightarrow U = 6.2eV - 4.6eV = 1.6eV$
Now $U = 1.6eV = e{V^{\prime}}$
So ${V^{\prime}} = 1.6V$
The maximum electric potential of the sphere came out to be 1.6eV. Option C is correct.
Note:Photoelectrons cannot be produced by light waves of any frequency. The photoelectric effect is brought on by light that is below a specific cut-off frequency. The more intense the light, above the cut-off voltage, the higher the kinetic energy the released photons would have. The quantity of photons released is inversely related to the frequency of incident light for a constant magnitude of light intensity.
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