
The ratio of slopes of maximum kinetic energy versus frequency and stopping potential (\[{{\text{V}}_{0}}\]) versus frequency, in photoelectric effect gives:
a.) Charge of electron
b.) Planck's constant
c.) Work function
d.) Threshold frequency
Answer
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Hint: To answer this question we should first know about photoelectric effect. To solve this question we should know about the equation of photoelectric effect.
Complete step by step solution:
Let us first know about the photoelectric effect. We should know that the photoelectric effect is a phenomenon in which electrons are ejected from the surface of a metal when light is incident on it. We call these ejected electrons as photoelectrons.
We should note that photoelectric effect occurs because the electrons at the surface of the metal tend to absorb energy from the incident light and use it to overcome the attractive forces that bind them to the metallic nuclei.
From this we can say that:
\[\text{hv}=\text{h}{{\text{v}}_{0}}+\text{e}{{\text{V}}_{0}}\]
Now, we can write this as:
\[\text{e}{{\text{V}}_{0}}=\text{hv}-\text{h}{{\text{v}}_{0}}\]
Now, we can write this as:
\[{{\text{V}}_{0}}=\dfrac{\text{h}}{\text{e}}\text{v}-\text{e}\dfrac{\text{h}}{{{\text{v}}_{0}}}~\]
y=mx+c
From this we can say that
\[{{\left( \text{Slope} \right)}_{1}}\] = $\dfrac{h}{e}$
Now, we will find the maximum kinetic energy:
${{(K.E)}_{\max }}=\text{hv}-\text{h}{{\text{v}}_{0}}~~~$
We will now find the slope of this equation:
The slope of this equation is:
\[{{\left( \text{Slope} \right)}_{2}}\] = h
So, we will now find the ratio of both slopes:
\[\dfrac{{{\left( Slope \right)}_{2}}}{{{\left( Slope \right)}_{1}}}=\dfrac{h}{\dfrac{h}{e}}=e\]
So, from this we can say that the ratio of slopes of maximum kinetic energy versus frequency and stopping potential (\[{{\text{V}}_{0}}\]) versus frequency is the charge of an electron. So, from this we can say that option A is correct.
Note: We should know that photoelectric cells are used for reproducing sound in cinematography. They are used for controlling the temperature of furnaces. Photoelectric cells are used for automatic switching on and off the street lights.
Complete step by step solution:
Let us first know about the photoelectric effect. We should know that the photoelectric effect is a phenomenon in which electrons are ejected from the surface of a metal when light is incident on it. We call these ejected electrons as photoelectrons.
We should note that photoelectric effect occurs because the electrons at the surface of the metal tend to absorb energy from the incident light and use it to overcome the attractive forces that bind them to the metallic nuclei.
From this we can say that:
\[\text{hv}=\text{h}{{\text{v}}_{0}}+\text{e}{{\text{V}}_{0}}\]
Now, we can write this as:
\[\text{e}{{\text{V}}_{0}}=\text{hv}-\text{h}{{\text{v}}_{0}}\]
Now, we can write this as:
\[{{\text{V}}_{0}}=\dfrac{\text{h}}{\text{e}}\text{v}-\text{e}\dfrac{\text{h}}{{{\text{v}}_{0}}}~\]
y=mx+c
From this we can say that
\[{{\left( \text{Slope} \right)}_{1}}\] = $\dfrac{h}{e}$
Now, we will find the maximum kinetic energy:
${{(K.E)}_{\max }}=\text{hv}-\text{h}{{\text{v}}_{0}}~~~$
We will now find the slope of this equation:
The slope of this equation is:
\[{{\left( \text{Slope} \right)}_{2}}\] = h
So, we will now find the ratio of both slopes:
\[\dfrac{{{\left( Slope \right)}_{2}}}{{{\left( Slope \right)}_{1}}}=\dfrac{h}{\dfrac{h}{e}}=e\]
So, from this we can say that the ratio of slopes of maximum kinetic energy versus frequency and stopping potential (\[{{\text{V}}_{0}}\]) versus frequency is the charge of an electron. So, from this we can say that option A is correct.
Note: We should know that photoelectric cells are used for reproducing sound in cinematography. They are used for controlling the temperature of furnaces. Photoelectric cells are used for automatic switching on and off the street lights.
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