In an experiment to study the photoelectric effect,the observed variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident radiation is as shown in the figure. The slope and $y$-intercept are respectively

When photons of energy $1 \ eV$ and $2.5 \ eV$ are incident on a metal surface with a work function of $0.5 \ eV$,what is the ratio of the maximum kinetic energies of the emitted photoelectrons?

The work-function of a metal is $1 eV$. Light of wavelength $3000 \text{Å}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

When photons are incident on a photosensitive material of work function $1.5 \text{ eV}$,the maximum velocity of the emitted photoelectrons is $8 \times 10^5 \text{ m/s}$. The stopping potential of the photoelectrons is (Mass of the electron $= 9 \times 10^{-31} \text{ kg}$ and charge of the electron $= 1.6 \times 10^{-19} \text{ C}$) (in $V$)

Statement $-1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{max}$ increase.
Statement $-2$: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

Light of frequency $4\nu_0$ is incident on a metal surface with a threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is: