When photons of energy $hv$ fall on a metal plate of work function $W_0$,photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of the radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be

$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)

If the frequency of incident radiation $(v)$ is increased,keeping other factors constant,the stopping potential ($v > v_0$,threshold frequency) will:

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

The maximum velocity of emitted photoelectrons is $4.8 \ m/s$. If the $e/m$ ratio of an electron is $1.76 \times 10^{11} \ C/kg$,then the stopping potential is ......

In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon: