Light of incident frequency $2$ times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{3}\right)^{\text{rd}}$ and intensity is doubled,then the photoelectric current will

When a photon with energy $8\ eV$ is incident on a metal surface having a threshold frequency of $1.6 \times 10^{15}\ Hz$,the maximum kinetic energy of the emitted photoelectrons is ............ $eV$. (Given: $h = 6.6 \times 10^{-34}\ J\cdot s$,$1\ eV = 1.6 \times 10^{-19}\ J$)

When a point source of monochromatic light is at a distance of $0.2 \ m$ from a photoelectric cell,the cut-off voltage and the saturation current are $0.6 \ V$ and $18 \ mA$ respectively. If the same source is placed $0.6 \ m$ away from the photoelectric cell,then:

When a light of wavelength $4900 Å$ falls on a photosensitive metal, a negative $2 \,V$ potential is required to stop the emitted electrons. Then, the work-function of the material is nearly (given charge on electron $= 1.602 \times 10^{-19} C$ and Planck's constant $= 6.625 \times 10^{-34} Js$) (in $eV$)

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

When light of wavelength $300 \; nm$ falls on a photoelectric emitter,photoelectrons are liberated. For another emitter,light of $600 \; nm$ wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters?