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:

When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W$), then the photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2}$, the stopping potential changes to $V_{2}$. Therefore, the charge of an electron $e$ is given by:

The work function of a metal is $2.3 \text{ eV}$ and the maximum kinetic energy of photoelectrons is $0.3 \text{ eV}$ when a photon is incident on it. Find the wavelength of the incident photon in $\mathring A$.

In the photoelectric experiment,if we use a monochromatic light,the $I - V$ curve is as shown. If the work function of the metal is $2 \, eV$,estimate the power of light used in $W$.
(Assume efficiency of photoemission $= 10^{-3} \%$,i.e.,the number of photoelectrons emitted is $10^{-3} \%$ of the number of photons incident on the metal.)

Let $v$ be the frequency of the incident photon and $v_0$ be the threshold frequency. The condition for the emission of photoelectrons is:

The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is: