The work functions of two photosensitive metal surfaces $A$ and $B$ are in the ratio $2: 3$. If $x$ and $y$ are the slopes of the graphs drawn between the stopping potential and frequency of incident light for the surfaces $A$ and $B$ respectively, then $x: y=$

Wave picture of light has failed to explain

The efficiency of a bulb of power 60 W is $16 \%$. The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of 2 m from the bulb is $\left(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right)$

The work function of a photosensitive metal surface is 1.1 eV . Two light beams of energies 1.5 eV and 2 eV incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is