When wavelength of incident photon is decrease then
A. Velocity of emitted photo-electrons decreased
B. Velocity of emitted photo-electrons increased
C. Velocity of emitted photo-electrons do not change
D. Photoelectric current increase

Hint: The total energy of a photon is equal to the sum of the energy utilized to eject an electron and the maximum kinetic energy of electrons. By using this concept kinetic energy can be obtained which further gives the velocity of the electron.

Formula used:
The energy of the photon is given by the equation is given as:
\[E = h\upsilon = \dfrac{{hc}}{\lambda }\].
Where \[h\] is the Plank constant, \[\upsilon \] is the frequency of incident light, c is the speed of light and \[\lambda \] is the wavelength.
Kinetic energy of photoelectrons is given as:
\[K{E_{\max }} = E - \phi \]
Where E is the energy and \[\phi \] is the work function.

Complete step by step solution:
As we know the energy of the photon is,
\[E = \dfrac{{hc}}{\lambda }\]
Here if the wavelength (\[\lambda \]) of the incident photon decreases, then energy will increase.
Kinetic energy of photoelectrons is,
\[KE = E - \phi \]
\[\Rightarrow \dfrac{1}{2}m{v^2} = E - \phi \]
In the above equation, work function(\[\phi \]) and mass (m) are constant. Hence if energy increases then kinetic energy will also be increased and thus velocity will increase. Therefore, when the wavelength of the incident photon decreased then the velocity of emitted photo-electrons increased.

Hence option B is the correct answer.

Note: The photoelectric is a phenomenon where electrons are ejected from a metal surface when light of sufficient frequency is incident on it. Einstein suggested that light behaved like a particle and that each particle of light has energy called a photon.