An image of the sun is formed by a lens of focal length $30 \ cm$ on the metal surface of a photoelectric cell and a photoelectric current $I$ is produced. The lens forming the image is then replaced by another of the same diameter but of focal length $15 \ cm$. The photoelectric current in this case is

$A$ metal surface having work function '$W_{0}$' emits photoelectrons when photons of energy '$E$' are incident on it. The electron enters a uniform magnetic field '$B$' in a perpendicular direction and moves in a circular path of radius '$r$'. Then '$r$' is equal to (where '$m$' and '$e$' are the mass and charge of the electron,respectively).

When a metal surface is illuminated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum velocities of photoelectrons ejected are $V$ and $2V$ respectively. The work function of the metal is ($h=$ Planck's constant,$c=$ velocity of light,$\lambda_1 > \lambda_2$).

Assertion : When ultraviolet light is incident on a photocell,its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X-$ rays,both $V_0$ and $K_{max}$ increase.
Reason : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

Electrons ejected from the surface of a metal,when light of a certain frequency is incident on it,are stopped fully by a retarding potential of $3 \ V$. The photoelectric effect on this metallic surface begins at a frequency of $6 \times 10^{14} \ s^{-1}$. The frequency of the incident light in $s^{-1}$ is: [Planck's constant $= 6.4 \times 10^{-34} \ J \cdot s$,charge on the electron $= 1.6 \times 10^{-19} \ C$]

$A$ photon of energy $8 eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} Hz$. The maximum kinetic energy of the emitted photoelectrons is ............... $eV$ (Given $h = 6.6 \times 10^{-34} Js$).