Assuming photoemission takes place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is:

$A$ metal surface is illuminated by light of a given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become

$A$ beam of photons with an energy of $10.5 \ eV$ strikes a metal plate. The photoelectrons are emitted with a maximum velocity of $1.6 \times 10^6 \ m \ s^{-1}$. The work function of the metal is (Assume mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$). (in $eV$)