The work functions of metal $A$ and $B$ are in the ratio $1: 2$. If light of frequency $f$ and $2f$ is incident on surface $A$ and $B$ respectively,then the ratio of the maximum kinetic energies of the emitted photoelectrons is:

Light of wavelength $\lambda_{\text{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. $A$ potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_e$,which of the following statement$(s)$ is(are) true?

On a photosensitive material,when the frequency of incident radiation is increased by $30 \%$,the kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.9 \ eV$. The work function of the surface is (in $eV$)

Light of wavelength $3500 \ \mathring A$ is incident on two metals $A$ and $B$. If their work functions are $4.2 \ eV$ and $1.9 \ eV$ respectively,which metal will emit photoelectrons?

The stopping potential of the photoelectrons from a photocell is:

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?