The ratio of work functions of two metals is $1:2$. If light of frequencies $f$ and $2f$ are incident on them respectively,what is the ratio of the maximum kinetic energy of the emitted photoelectrons?

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then,the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively,is

The work function of a metal surface is $1.6 \times 10^{-19} \text{ J}$. When light of wavelength $6400 \ \mathring{A}$ is incident on this surface,the maximum kinetic energy of the emitted photoelectrons is ....... $\text{J}$.

When a metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for photoelectric current is $V_0$. When the same surface is illuminated by light of frequency $\frac{v}{2}$,the stopping potential is $\frac{V_0}{4}$,the threshold frequency of photoelectric emission is

If the frequency of incident radiation $(v)$ is increased,keeping other factors constant,the stopping potential ($v > v_0$,threshold frequency) will: