The frequency of incident light falling on a photosensitive material is doubled. The kinetic energy $(K.E.)$ of the emitted photoelectrons will be:

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

If the maximum velocity with which an electron can be emitted from a photocell is $4 \times 10^8 \, cm/s$,the stopping potential is ................ $V$ (mass of electron $= 9 \times 10^{-31} \, kg$).

When a light of wavelength '$\lambda$' falls on the emitter of a photosensitive surface, the maximum speed of emitted photoelectrons is '$V$'. If the incident wavelength is changed to '$2\lambda/3$',the maximum speed of emitted photoelectrons will be

Light of wavelength $\lambda$ falls on a metal having work function $\frac{hc}{\lambda_0}$. Photoelectric effect will take place only if ($\lambda_0$ is the threshold wavelength).

If the light from the Balmer series of hydrogen is used to eject photoelectrons from a metal, then the maximum work function of the metal can be: (in $\text{ eV}$)