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# Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required for these electrons areA. $5.5\,V$B. $1.5\,V$C. $9.5\,V$D. $4.0\,V$

Last updated date: 12th Aug 2024
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Hint: The maximum kinetic energy of the emitted photoelectrons is the extra energy of the photon which remains after getting consumed part of it in overcoming the work function of the metal. Using conversion of kinetic energy in work done to stop by applying potential difference, we can get 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 minimum energy needed to overcome the attractive potential of the valence electron of the atom and eject it is called the work function of the metal. The wavelength of the photon corresponding to the work function of the metal is called the threshold wavelength. As the energy of the photon is inversely proportional to the wavelength of the photon, so for minimum energy the wavelength should be maximum.

The stopping potential is the potential which is able to stop the ejected electron. Using the work-energy theorem, when a body is acted upon by external force then the change in kinetic energy of the body is equal to the work done on it.

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 maximum kinetic energy of the ejected electron is given as 4.0eV.
$e{V_0} = 4eV$
$\therefore {V_0} = 4.0\,V$
Hence, the stopping potential is 4.0 V.

Therefore, the correct option is D.

Note: We should be careful about the data given. The stopping potential is equal to the potential which is needed to be applied to stop a moving electron. So, the stopping potential should be corresponding to the kinetic energy of the electron, not the energy of the photon.