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(A) According to the photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$. Here,$V_s$ is the stopping potential and $\phi$ is the work function.

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increase

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(A) According to the photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$. Here,$V_s$ is the stopping potential and $\phi$ is the work function. The kinetic energy of the emitted photoelectrons is given by $K_{max} = eV_s = \frac{1}{2}mv^2$. If the velocity $v$ of the emitted photoelectrons is kept constant,the kinetic energy $K_{max}$ remains constant. However,as the incident wavelength $\lambda$ is decreased,the energy of the incident photon $E = \frac{hc}{\lambda}$ increases. Since $E = \phi + K_{max}$,and $K_{max}$ is constant while $E$ increases,the work function $\phi$ of the material would effectively need to change or the stopping potential $V_s$ must increase to balance the equation if we consider the energy conservation principle for a fixed material. Actually,for a fixed material,if $\lambda$ decreases,$K_{max}$ increases. If the problem states the velocity is kept constant,it implies an external adjustment or a misunderstanding of the physical constraints; however,based on the standard photoelectric equation $\frac{hc}{\lambda} = \phi + eV_s$,if $\lambda$ decreases,the energy of the incident photon increases,which leads to an increase in the stopping potential $V_s$.

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

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