$A$ photon released by the transition of an electron from the second excited state to the ground state of a Hydrogen atom is incident on the surface of a metal with a work function of $3.1 \ eV$. The de Broglie wavelength of the most energetic electron emitted from that metal surface is nearly:

The work functions for metals $A$, $B$, and $C$ are $1.92 eV$, $2.0 eV$, and $5 eV$ respectively. The metal(s) which will emit photoelectrons for incident radiation of wavelength $4100 Å$ is/are $[h=6.63 \times 10^{-34} J s, e=1.6 \times 10^{-19} C, c=3 \times 10^8 m/s]$.

$A$ 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:

According to Einstein's photoelectric equation,the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

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.

Why maximum kinetic energy of a photoelectron cannot be negative?