The retarding potential necessary to stop the emission of photoelectrons,when a target material of work function $1.24 eV$ is irradiated with light of wavelength $4.36 \times 10^{-7} m$ is (in $eV$)

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 .............. $V$.

Light of two different frequencies whose photons have energies $1.3 eV$ and $2.8 eV$ respectively,successfully illuminate a metallic surface whose work function is $0.8 eV$. The ratio of maximum speeds of emitted electrons will be

The magnetic field associated with a light wave is given,at the origin,by $B = B_0 [\sin(3.14 \times 10^7 ct) + \sin(6.28 \times 10^7 ct)]$. If this light falls on a silver plate having a work function of $4.7 \ eV$,what will be the maximum kinetic energy of the photoelectrons? (in $eV$)

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photo-electrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then: