
The light rays having photons of energy 1.8 eV are falling on metal surface having a work function 1.2 eV. What is the stopping potential to be applied in order to stop the emitting electrons?
A. 3V
B. 1.2V
C. 0.6V
D. 1.4V
Answer
162k+ views
Hint: To find the stopping potential, we can use the work-energy theorem which states that the net work done is equal to the change in kinetic energy. So, the work done by the applied potential is such that the final kinetic energy of the photoelectron becomes zero. That potential is called the stopping potential.
Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[\phi \] is the work function of the metal.
Complete step by step solution:
The kinetic energy of the photo-electron is the remaining energy of the photon transferred to the ejected electron in the form of kinetic energy. So, the kinetic energy of the photoelectron is calculated by finding the difference of total energy of the photon and the work function of the metal.
When the photoelectron stops on application of the potential difference, then the magnitude of the change in kinetic energy is equal to the maximum kinetic energy of the photoelectron. So, the work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron.
\[e{V_0} = K{E_{\max }}\]
The energy of the photon is given as 1.8 eV and the work function of the metal surface is 1.2 eV. The maximum kinetic energy of the ejected electron will be,
\[K = 1.8eV - 1.2eV\]
\[\Rightarrow K = 0.6\,eV\]
So, the stopping potential will be,
\[e{V_0} = 0.6\,eV\]
\[\therefore {V_0} = 0.6V\]
Hence, the stopping potential is 0.6V.
Therefore, the correct option is C.
Note: While using the work-energy theorem, we don’t need to put the negative sign in the value of the change in kinetic energy of the photoelectron. So, the stopping potential energy is equal to the kinetic energy of the photoelectron.
Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[\phi \] is the work function of the metal.
Complete step by step solution:
The kinetic energy of the photo-electron is the remaining energy of the photon transferred to the ejected electron in the form of kinetic energy. So, the kinetic energy of the photoelectron is calculated by finding the difference of total energy of the photon and the work function of the metal.
When the photoelectron stops on application of the potential difference, then the magnitude of the change in kinetic energy is equal to the maximum kinetic energy of the photoelectron. So, the work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron.
\[e{V_0} = K{E_{\max }}\]
The energy of the photon is given as 1.8 eV and the work function of the metal surface is 1.2 eV. The maximum kinetic energy of the ejected electron will be,
\[K = 1.8eV - 1.2eV\]
\[\Rightarrow K = 0.6\,eV\]
So, the stopping potential will be,
\[e{V_0} = 0.6\,eV\]
\[\therefore {V_0} = 0.6V\]
Hence, the stopping potential is 0.6V.
Therefore, the correct option is C.
Note: While using the work-energy theorem, we don’t need to put the negative sign in the value of the change in kinetic energy of the photoelectron. So, the stopping potential energy is equal to the kinetic energy of the photoelectron.
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