According to Einstein's photoelectric equation,the graph of the kinetic energy of photoelectrons emitted from a metal versus the frequency of incident radiation is a straight line whose slope:

Why maximum kinetic energy of a photoelectron cannot be negative?

Light of frequency $7.21 \times 10^{14} \; Hz$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5} \; m/s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

$A$ photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ (where $\lambda < \frac{hc}{\phi_0}$). The fastest photoelectron has a de Broglie wavelength $\lambda_d$. $A$ change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_d$ in $\lambda_d$. Then the ratio $\frac{\Delta \lambda_d}{\Delta \lambda}$ is proportional to:

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.

The work function of a photosensitive metal surface is $\phi$. When a photon of energy $3\phi$ is incident on the surface,a photoelectron with a maximum velocity of $6.6 \times 10^6 \ m/s$ is emitted. If the energy of the incident photon is increased to $9\phi$,the maximum velocity of the emitted photoelectron will be ..........