Calculate the stopping potential if the maximum kinetic energy of the photoelectrons emitted by a photocell will be $3eV$.
Answer
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Hint: The maximum kinetic energy of the photoelectrons emitted by the photocell is defined by taking the product of the electronic charge and the stopping potential. Rearrange the equation in terms of the stopping potential and substitute this in the equation. Hope these all may help you to solve this question.
Complete answer:
First of all let us mention what all are given in the question. The maximum kinetic energy of the photoelectrons emitted by a photocell will be given as,
$E=3eV$
The electronic charge is given by a value of,
$e=1.6\times {{10}^{-19}}C$
The maximum kinetic energy of the photoelectron is given by the equation,
$E=e{{V}_{0}}$
Where ${{V}_{0}}$be the stopping potential.
When we substitute the value of kinetic energy in this equation, the energy should be in unit of joules. Then only the correct value of potential is obtained. So let us convert the kinetic energy in electron volts to joules.
The kinetic energy in electron volt is given as,
$E=3eV$
It can be expressed in joules by the multiplication of electronic charge with it. That is,
\[E=3\times 1.6\times {{10}^{-19}}J\]
Now let us substitute all the values obtained in the equation of kinetic energy.
$\begin{align}
& E=e{{V}_{0}} \\
& \Rightarrow 3\times 1.6\times {{10}^{-19}}J=1.6\times {{10}^{-19}}\times {{V}_{0}} \\
\end{align}$
Rearranging the equation will give,
$\dfrac{3\times 1.6\times {{10}^{-19}}J}{1.6\times {{10}^{-19}}}={{V}_{0}}$
Simplifying the equation,
\[{{V}_{0}}=3V\]
Note:
The stopping voltage otherwise called as stopping potential is defined as the voltage difference needed to prevent the electrons from moving between the plates and creating a current in the photoelectric experiment. The stopping potential has been found to be linearly varying with frequency of incident light. If there is an increase in the frequency of the incident light, then the kinetic energy of the emitted electrons also increases. Therefore greater retarding potential will be needed to stop them completely.
Complete answer:
First of all let us mention what all are given in the question. The maximum kinetic energy of the photoelectrons emitted by a photocell will be given as,
$E=3eV$
The electronic charge is given by a value of,
$e=1.6\times {{10}^{-19}}C$
The maximum kinetic energy of the photoelectron is given by the equation,
$E=e{{V}_{0}}$
Where ${{V}_{0}}$be the stopping potential.
When we substitute the value of kinetic energy in this equation, the energy should be in unit of joules. Then only the correct value of potential is obtained. So let us convert the kinetic energy in electron volts to joules.
The kinetic energy in electron volt is given as,
$E=3eV$
It can be expressed in joules by the multiplication of electronic charge with it. That is,
\[E=3\times 1.6\times {{10}^{-19}}J\]
Now let us substitute all the values obtained in the equation of kinetic energy.
$\begin{align}
& E=e{{V}_{0}} \\
& \Rightarrow 3\times 1.6\times {{10}^{-19}}J=1.6\times {{10}^{-19}}\times {{V}_{0}} \\
\end{align}$
Rearranging the equation will give,
$\dfrac{3\times 1.6\times {{10}^{-19}}J}{1.6\times {{10}^{-19}}}={{V}_{0}}$
Simplifying the equation,
\[{{V}_{0}}=3V\]
Note:
The stopping voltage otherwise called as stopping potential is defined as the voltage difference needed to prevent the electrons from moving between the plates and creating a current in the photoelectric experiment. The stopping potential has been found to be linearly varying with frequency of incident light. If there is an increase in the frequency of the incident light, then the kinetic energy of the emitted electrons also increases. Therefore greater retarding potential will be needed to stop them completely.
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