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

Energy of the incident photons on the photosensitive metal surface is $3W$ and then $5W$ where '$W$' is the work function of that metal. The ratio of velocities of the emitted photoelectrons is

This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1:$ $A$ metallic surface is irradiated by a monochromatic light of frequency $v > v_0$ (the threshold frequency). If the incident frequency is now doubled,the photocurrent and the maximum kinetic energy are also doubled.
Statement $2:$ The maximum kinetic energy of photoelectrons emitted from a surface is linearly dependent on the frequency of the incident light. The photocurrent depends only on the intensity of the incident light.

The work function of a metal is $6.825 \ eV$. Its threshold wavelength is approximately: (Given $c = 3 \times 10^8 \ m/s$)

When ultraviolet radiation of a certain frequency falls on a potassium target,the photoelectrons released can be stopped completely by a retarding potential of $0.6 \, V$. If the frequency of the radiation is increased by $10 \%$,this stopping potential rises to $0.9 \, V$. The work function of potassium is ........ $eV$.

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