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 maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

When a light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are emitted. For another emitter,light of wavelength $600 \ nm$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

Photons with energy $5\, eV$ are incident on a cathode $C$ of a photoelectric cell. The maximum kinetic energy of the emitted photoelectrons is $2\, eV$. When photons of energy $6\, eV$ are incident on $C$,no photoelectrons will reach the anode $A$ if the stopping potential of $A$ relative to $C$ is .............. $V$.

The threshold frequency of a metal with work function $6.63 \ eV$ is: