When light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of the emitted photoelectrons is $V$. If the incident wavelength is changed to $\frac{2\lambda}{3}$,the maximum speed of the emitted photoelectrons will be:

Two monochromatic light beams $A$ and $B$ of equal intensity are incident normally on a metallic surface per unit area. Their wavelengths are $\lambda_A$ and $\lambda_B$ respectively. The incident light is effective in ejecting photoelectrons. The ratio of the number of photoelectrons emitted by beam $A$ to that by beam $B$ is:

Photoelectric emission is observed from a metallic surface for frequencies $\nu_1$ and $\nu_2$ of the incident light rays $(\nu_1 > \nu_2)$. If the ratio of the maximum kinetic energy of the photoelectrons emitted in the first case to that in the second case is $3 : K$, then the threshold frequency of the metallic surface is:

The work function of caesium metal is $2.14 \; eV$. When light of frequency $6 \times 10^{14} \; Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
$(a)$ maximum kinetic energy of the emitted electrons,
$(b)$ stopping potential,and
$(c)$ maximum speed of the emitted photoelectrons?

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

If the stopping potential of photoelectrons emitted by a source placed at a distance of $0.2 \ m$ from a photocell is $0.6 \ V$,then the stopping potential of photoelectrons emitted by the same source placed at a distance of $0.6 \ m$ from the photocell will be .......... $V$.