If a metal sheet is irradiated with radiations of frequencies $v_1$ and $v_2$ and the kinetic energies of the emitted photoelectrons are in the ratio $1 : x$, then the threshold frequency of the metal is:

In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

When light of frequency twice the threshold frequency is incident on a metal plate,the maximum velocity of the emitted electron is $v_{1}$. When the frequency of incident radiation is increased to five times the threshold value,the maximum velocity of the emitted electron becomes $v_{2}$. If $v_{2} = x v_{1}$,the value of $x$ will be...........

According to the photoelectric effect,the plot of kinetic energy of the emitted photo-electrons from a metal versus the frequency of the incident radiation gives a straight line whose slope

Photons of wavelength $\lambda$ emitted by a source of power $P$ are incident on a photocell. If the current produced in the cell is $I$,then the percentage of incident photons which produce current in the photocell is: (where $h$ is Planck's constant and $c$ is the speed of light in vacuum)

Light of two different frequencies with photon energies of $1 \ eV$ and $2.5 \ eV$ are incident one after another on a metal surface having a work function of $0.5 \ eV$. The ratio of the maximum kinetic energy of the emitted photoelectrons will be: