The photoelectric threshold frequency of a metal is $v$. When light of frequency $4v$ is incident on the metal,the maximum kinetic energy of the emitted photoelectrons is

The work function of a metal is $1.6 \times 10^{-19} \, J$. When the metal surface is illuminated by light of wavelength $6400 \, \mathring{A}$,the maximum kinetic energy of the emitted photo-electrons will be. (Planck's constant $h = 6.4 \times 10^{-34} \, Js$)

When a photon of energy $hv$ is incident on an aluminum plate (work function $E_0$),photoelectrons with maximum kinetic energy $K$ are emitted. What will be the maximum kinetic energy of the emitted photoelectrons when a photon of energy $2hv$ is incident on the same aluminum plate?

Radiations of two photons having energies twice and five times the work function of a metal are incident successively on a metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will be

The slope of the graph between stopping potential and the frequency of incident radiation for the photoelectric effect is . . . . . . .

In a photoelectric experiment, the slope of the graph drawn between stopping potential $(V_s)$ along the $y$-axis and the frequency $(\nu)$ of incident radiation along the $x$-axis is (Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$)