Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

$A$ beam of light of wavelength $\lambda$ falls on a metal having work function $\phi$ placed in a magnetic field $B$. The most energetic electrons, moving perpendicular to the field, are bent in circular arcs of radius $R$. If the experiment is performed for different values of $\lambda$, then the $B^2$ vs. $\frac{1}{\lambda}$ graph will look like (keeping all other quantities constant):

When photons of energy $h \nu$ fall on a photosensitive surface of work function $E_0$,photoelectrons of maximum kinetic energy $k$ are emitted. If the frequency of radiation is doubled,the maximum kinetic energy will be equal to ($h=$ Planck's constant).

$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

When a light of wavelength $4900 Å$ falls on a photosensitive metal, a negative $2 \,V$ potential is required to stop the emitted electrons. Then, the work-function of the material is nearly (given charge on electron $= 1.602 \times 10^{-19} C$ and Planck's constant $= 6.625 \times 10^{-34} Js$) (in $eV$)

According to Einstein's photoelectric equation,the graph between the kinetic energy of emitted photoelectrons and the frequency of incident radiation is: