$A$ 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:

In the experiment of photoelectric effect, $K_{max}$ for a photoelectron is $K_0$. If the frequency of light is doubled, then $K_{max}$ is $3K_0$. What would be the value of $K_{max}$ if the frequency of light is tripled? (in terms of $K_0$)

Two radiations of photon energies $1 \; eV$ and $2.5 \; eV$ successively illuminate a photosensitive metallic surface of work function $0.5 \; eV$. The ratio of the maximum speeds of the emitted electrons is

The electric field at the point associated with a light wave is given by $E = 200 [\sin(6 \times 10^{15} t) + \sin(9 \times 10^{15} t)] \, Vm^{-1}$. Given $h = 4.14 \times 10^{-15} \, eVs$. If this light falls on a metal surface having a work function of $2.50 \, eV$,the maximum kinetic energy of the photoelectrons will be ........... $eV$.

The surface of a metal is illuminated alternately with photons of energies $E_{1} = 4 \ eV$ and $E_{2} = 2.5 \ eV$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal in $eV$ is:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
$Assertion$ $A$ : Number of photons increases with increase in frequency of light.
$Reason$ $R$ : Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
In the light of the above statements,choose the most appropriate answer from the options given below :