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

Radiation of wavelength $332 \ nm$ is incident on a metal surface having a work function of $1.07 \ eV$. The stopping potential required to stop the emission of photoelectrons from the metal surface is ............ $V$. $(h = 6.6 \times 10^{-34} \ J s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

In a photoelectric experiment for $4000 \, \mathring{A}$ incident radiation,the potential difference to stop the ejection is $2 \, V$. If the incident light is changed to $3000 \, \mathring{A}$,then the potential required to stop the ejection of electrons will be

$A$ beam of electromagnetic radiation of intensity $6.4 \times 10^{-5} \; W/cm^{2}$ is comprised of wavelength $\lambda = 310 \; nm$. It falls normally on a metal surface (work function $\varphi = 2 \; eV$) of surface area $1 \; cm^{2}$. If one in $10^{3}$ photons ejects an electron, the total number of electrons ejected in $1 \; s$ is $10^{x}$. Then $x$ is: $(hc = 1240 \; eV \cdot nm, 1 \; eV = 1.6 \times 10^{-19} \; J)$

Which one of the following graphs correctly represents the variation of maximum kinetic energy $(E_{k})$ of the emitted electrons with frequency $(\nu)$ of incident light in the photoelectric effect?

Which one of the four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident radiation $(I)$ is correct?