The maximum kinetic energy of emitted photoelectrons depends on

When photons of energy $hf$ fall on an aluminium plate (of work function $E_0$),photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be:

The stopping potential required to reduce the photoelectric current to zero is . . . . . .

In an experiment on photoelectric emission from a metallic surface,the wavelength of incident light is $2 \times 10^{-7} \,m$ and the stopping potential is $2.5 \,V$. The threshold frequency of the metal (in $Hz$) is approximately (charge of electron $e=1.6 \times 10^{-19} \,C$,Planck's constant $h=6.6 \times 10^{-34} \,J-s$):

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$,when light of frequency $\nu$ is incident on it,is $\lambda$. Then $\nu =$ (where $h$ is Planck's constant and $m$ is the mass of the electron).

In the photoelectric emission process from a metal with a work function of $1.8\, eV$,the kinetic energy of the most energetic electrons is $0.5\, eV$. The corresponding stopping potential is ......... $V$.