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

The threshold frequency of a metal is $f_0$. When light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of the photoelectrons is $v_1$. When the frequency of the incident radiation is increased to $5f_0$, the maximum velocity of the photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is

In the experiment of $P.E.E.$,the $KE_{max}$ of an electron is $K_0$. If the frequency is increased by a factor of $n_1$,then the $KE_{max}$ becomes $n_2K_0$. Find the work function.

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?

When light of frequency twice the threshold frequency is incident on a metal plate,the maximum velocity of the emitted electron is $v_{1}$. When the frequency of incident radiation is increased to five times the threshold value,the maximum velocity of the emitted electron becomes $v_{2}$. If $v_{2} = x v_{1}$,the value of $x$ will be...........

The photoelectric threshold frequency of a metal is $v$. When light of frequency $4v$ is incident on the metal,the maximum kinetic energy of the emitted photoelectrons is