The light of two different frequencies whose photons have energies $3.8 \, eV$ and $1.4 \, eV$ respectively,illuminate a metallic surface whose work function is $0.6 \, eV$ successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be

If light of wavelength $6600 \ \mathring A$ is incident on a metal surface with a work function of $2 \ eV$,what will be the maximum kinetic energy of the emitted photoelectrons?

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?

The work function of a metal is $h \nu_0$. Light of frequency $\nu$ falls on this metal. The photoelectric effect will take place only if

The maximum velocity of electrons emitted from a metal surface is $V$,when the frequency of light falling on it is $f$. What is the maximum velocity when the frequency becomes $4f$?

When the electromagnetic radiations of frequencies $4 \times 10^{15} \,Hz$ and $6 \times 10^{15} \,Hz$ fall on the same metal in different experiments,the ratio of maximum kinetic energy of electrons liberated is $1: 3$. The threshold frequency for the metal is ............... $\times 10^{15} Hz$.