Answer
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Hint: To answer this question we first need to understand the term stopping potential which is also known as cutoff voltage or cutoff potential. The cutoff or stopping potential is the required voltage or potential to stop the removal of the electrons from the given metal surface when the incident energy of the radiation is larger than the work potential of the metal on which the radiation is incident.
Complete step by step answer:
The stopping potential is generally observed in the photoelectric effect. In the photoelectric effect when light or radiation incident on any metal surface the electrons absorb the energy from the photons and tend to move as a result in the form of an electric current.
The formula of stopping potential can be given as,
$V{\text{ }} = {\text{ }}\dfrac{{\left( {hf{\text{ }} - {\text{ }}\phi } \right)}}{q}{\text{ }}$
where $\phi $ is the work function defined as the amount of energy needed to bind the electrons in the metals, $hf$ is the energy of photons, while $q$ is the charge.
Basically, the stopping potential can be considered as a negative potential at which the photoelectric currents become zero. The stopping potential does not depend on the intensity nor the number of incident photons but the stopping potential depends on the frequency of the incident light, the higher the frequency of the incident light higher the stopping potential or cut potential. It also depends on the kinetic energy of the electrons.
Hence, the correct answer is option (A).
Note: The stopping potential or known as cut off potential is mainly used to determine the kinetic energy that the electrons carry when they are ejected from the metal plate. The product of the charge $q$ on an electron and the stopping voltage $V$ provides the maximum kinetic energy of the electrons that are ejected from the metal plate.
Complete step by step answer:
The stopping potential is generally observed in the photoelectric effect. In the photoelectric effect when light or radiation incident on any metal surface the electrons absorb the energy from the photons and tend to move as a result in the form of an electric current.
The formula of stopping potential can be given as,
$V{\text{ }} = {\text{ }}\dfrac{{\left( {hf{\text{ }} - {\text{ }}\phi } \right)}}{q}{\text{ }}$
where $\phi $ is the work function defined as the amount of energy needed to bind the electrons in the metals, $hf$ is the energy of photons, while $q$ is the charge.
Basically, the stopping potential can be considered as a negative potential at which the photoelectric currents become zero. The stopping potential does not depend on the intensity nor the number of incident photons but the stopping potential depends on the frequency of the incident light, the higher the frequency of the incident light higher the stopping potential or cut potential. It also depends on the kinetic energy of the electrons.
Hence, the correct answer is option (A).
Note: The stopping potential or known as cut off potential is mainly used to determine the kinetic energy that the electrons carry when they are ejected from the metal plate. The product of the charge $q$ on an electron and the stopping voltage $V$ provides the maximum kinetic energy of the electrons that are ejected from the metal plate.
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