The stopping potential for a photoelectric emission process is $10 \ V$. The maximum kinetic energy of the electrons ejected in the process is [Charge on electron $e = 1.6 \times 10^{-19} \ C$]

Maximum velocity of photoelectrons emitted by a metal surface is $1.2 \times 10^6\, m/s$. Assuming the specific charge of the electron to be $1.8 \times 10^{11}\, C/kg$,the value of the stopping potential in volt will be

When a point source of monochromatic light is at a distance of $0.2 \ m$ from a photoelectric cell,the stopping potential and saturation photocurrent are $0.6 \ V$ and $18 \ mA$ respectively. If the same source is placed $0.6 \ m$ away from the photoelectric cell,then:

The stopping potential $(V_0)$ versus frequency $(\nu)$ plot of a substance is shown in the figure. The threshold wavelength is:

$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

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