The photoelectric threshold wavelength for a certain metal surface is $3600 \mathring A$. If the metal surface is irradiated by a wavelength of $1100 \mathring A$,the kinetic energy of the emitted photoelectrons is (in $\text{ eV}$)

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

If the work function of a photometal is $6.825 \; eV$,its threshold wavelength will be ............. $\mathring A$ $(c = 3 \times 10^8 \; m/s)$.

Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass $m$ are $V_1$ and $V_2$ respectively,then ($h=$ Planck's constant):

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is [$h=$ Planck's constant,$m=$ mass of electron and $c=$ speed of light]